How can I see CakePHP's SQL dump in the controller? - sql

Is there a way that one can cause CakePHP to dump its SQL log on demand? I'd like to execute code up until a point in my controller and see what SQL has been run.

Try this:
$log = $this->Model->getDataSource()->getLog(false, false);
debug($log);
http://api.cakephp.org/2.3/class-Model.html#_getDataSource
You will have to do this for each datasource if you have more than one though.

There are four ways to show queries:
This will show the last query executed of user model:
debug($this->User->lastQuery());
This will show all executed query of user model:
$log = $this->User->getDataSource()->getLog(false, false);
debug($log);
This will show a log of all queries:
$db =& ConnectionManager::getDataSource('default');
$db->showLog();
If you want to show all queries log all over the application you can use in view/element/filename.ctp.
<?php echo $this->element('sql_dump'); ?>

If you're using CakePHP 1.3, you can put this in your views to output the SQL:
<?php echo $this->element('sql_dump'); ?>
So you could create a view called 'sql', containing only the line above, and then call this in your controller whenever you want to see it:
$this->render('sql');
(Also remember to set your debug level to at least 2 in app/config/core.php)
Source

for cakephp 2.0
Write this function in AppModel.php
function getLastQuery()
{
$dbo = $this->getDatasource();
$logs = $dbo->getLog();
$lastLog = end($logs['log']);
return $lastLog['query'];
}
To use this in Controller
Write : echo $this->YourModelName->getLastQuery();

It is greatly frustrating that CakePHP does not have a $this->Model->lastQuery();. Here are two solutions including a modified version of Handsofaten's:
1. Create a Last Query Function
To print the last query run, in your /app_model.php file add:
function lastQuery(){
$dbo = $this->getDatasource();
$logs = $dbo->_queriesLog;
// return the first element of the last array (i.e. the last query)
return current(end($logs));
}
Then to print output you can run:
debug($this->lastQuery()); // in model
OR
debug($this->Model->lastQuery()); // in controller
2. Render the SQL View (Not avail within model)
To print out all queries run in a given page request, in your controller (or component, etc) run:
$this->render('sql');
It will likely throw a missing view error, but this is better than no access to recent queries!
(As Handsofaten said, there is the /elements/sql_dump.ctp in cake/libs/view/elements/, but I was able to do the above without creating the sql.ctp view. Can anyone explain that?)

In CakePHP 1.2 ..
$db =& ConnectionManager::getDataSource('default');
$db->showLog();

What worked finally for me and also compatible with 2.0 is to add in my layout (or in model)
<?php echo $this->element('sql_dump');?>
It is also depending on debug variable setted into Config/core.php

Plugin DebugKit for cake will do the job as well. https://github.com/cakephp/debug_kit

If you are interested in some specific part of code, you can clear first the log, and then display only queries that happen after that point.
Also note that 'Model' below, is the actual class name, like User, Page etc.
//clear log (boolean $clear = true)
$this->Model->getDataSource()->getLog(false, true);
...
...
...
...
//Show log so far
$log = $this->Model->getDataSource()->getLog(false, false);
debug($log);
exit;

Related

passing msqli to a function - can't suss out why it's not working

I've searched high and low for an answer on this, but I'm either missing something, or I just can't find anything in this context.
Background - trying to avoid spaghetti frenzy with a little casual project I'm starting; part of this will involve mainly just calls to a mysql database, displaying table content and so on. Simply put, like a CRM I guess.
I may be way off base here, but I want to be able to create my sql calls as functions, which will help if/when I tweak and tune, as well as creating a cleaner code for what I'm looking to do.
So, without further ado, I have this as a demomstration:
echo "<table>";
selectall('actions','content',$mysqli);
echo "</table><br><br>";
What this does is show all rows from my table of 'actions. "content" is just an example field name I'm passing through that I want to display, as it is the main human-relevant field name in that table. I'm also passing $mysqli through here for my function db call.
My function looks like this:
function selectall($s_table,$s_content,$mysqli){
$query = "SELECT * FROM " . $s_table;
$resource = $mysqli->query($query);
if ( !$resource ) throw new Exception($db->error);
while ( $row = $resource->fetch_assoc() ) {
$id = $row['id'];
echo "<tr><td>{$row[$s_content]}</td></tr>";
}
$resource->free();
$mysqli->close();
}
However.... it doesn't work, and it seems to throw a wobbly saying:
Warning: mysqli::query(): Couldn't fetch mysqli
This points to the action within the line $resource = $mysqli->query($query);
I know the function and everything is ok, as if I restate and declare $mysqli within the first line of the function, like so...
$mysqli = new mysqli(username password and so on in here);
... it works spot on.
$mysqli exists and works within the same code that is passing the variable within the function too.
This is early stages, so by shuffling the code around trying to poke the $mysqli pass into life I have perhaps made the code a little messier that intended, so try not to worry too much about that.
Anyone any ideas why it doesn't like this?
D'oh...
I had a
$mysqli->close();
in the lines above. Solved myself.
For reference, this is my function:
function selectall($s_table,$s_field,$mysqli){
if ($mysqli->connect_error) {die('Connect Error (' . $mysqli->connect_errno . ') '. $mysqli->connect_error);}
$s_table = preg_replace('/[^0-9a-zA-Z_]/', '', $s_table); // Cleans up the table name variable
$s_field = preg_replace('/[^0-9a-zA-Z_]/', '', $s_field); // Cleans up the field name variable
$query = "SELECT * FROM " . $s_table; // Adds passed table name to the select all
$resource = $mysqli->query($query);
if ( !$resource ) throw new Exception($db->error);
while ( $row = $resource->fetch_assoc() ) {
echo "<tr><td>{$row[$s_field]}</td></tr>"; // Content for each row of the select all
}
$resource->free();
$mysqli->close();
}
As you can see, I've also tried to protect the variables that enter the function.
This can be called via:
selectall('actions','content',$mysqli);
In this context, I want to view all the entries in the 'actions' table by the field name 'content'. This function, with some code above and below for a table, will create a new row for each entry.
I'll probably evolve a few, already created on that includes a delete button at the end of the line which is 'selectalldel'.
Open to comments on whether this actually is worthwhile, but thought I'd post up my corrected stupidity in case anyone finds this useful.

Possibly missing something very simple in this PHP foreach loop

This question is an extension for this question.
Basically the script compares two arrays and outputs the difference (what to show to the user).
$urlsToShow = array_diff($siteUrls, $seenUrls);
if (!empty($urlsToShow)) {
// Echo the url to show for the iframe within browse.php and add an entry to the database that the user has seen this site
foreach ($urlsToShow as $urlToShow) {
echo $urlToShow;
$entry = "INSERT INTO views VALUES ('', '$currentUsername', '$urlToShow')";
mysqli_query($con, $entry);
break;
}
}
The problem is that I get two entries into the database with one iteration? The first site from the $urlsToShow array is displayed (echoed into the iframe) and gets added to the database along with the next site from the same array. But the user will never see the next site.
Am I using break incorrectly?
It is possible that using break; will not be in your best interest. I suggest using continue;.

can there be a tooltip with data from the database

I need to know can there be a tooltip populated with data from database to be displayed in the tooltip.
something like the tooltip should contain
name stauts
abc active
xyz active
pqr active
name and status are retrived from db
I need this tooltip onmouseover, am using CJSON decoded to render the content
i did go google but hardly did find that i would throughly understand and implement.
can anyone out there has any ideas for what am looking.
There is a extension named yii-bootstrap, which described clearly here.
For using tooltip easily in this extension, just look here.
I use cluetip for this. Its not related to Yii but will give you some idea :
JS
function renderInfoTips(opts){
var elements=$('#'+opts.form).get(0).elements;
for(i=0; i<opts.tips.length;i++){
$(elements[opts.tips[i].field]).parent().prepend(opts.tips[i].tip);
}
var clue_opts={arrows:true,splitTitle: '|',closePosition: 'title',sticky:true,dropShadow:false,mouseOutClose:true,
onShow:function(ct, ci){
if(!$.browser.webkit) $(ct).css('top',$(ct).position().top- 30+'px');
}
}
$('#'+opts.form).find(".infotip").cluetip(clue_opts);
}
PHP
function setInfoTipsJavascript($form_id,infotips){
if (count($this->infotips) <1 ) return '';
//get all tip names
$names_csv=join(',',array_keys(infotips));
//get tips details from db
$query="select name, description from infotips where FIND_IN_SET(name ,'$names_csv')";
//run the query, in Yii you have to use InfoTipsModel , I have skipped that portion
//$infotipS , lets say this is query object
$tips=array();
while($tip=$infotipS->Assoc()){
$this->infotips[$tip['name']]['tip']="<a href='javascript:void(0)' class='infotip' title='|{$tip['description']}'> </a>";
$tips[]=$this->infotips[$tip['name']];
}
$tips=json_encode($tips);
$script="\nrenderInfoTips({\"form\":'{$form_id}', \"tips\":{$tips}});\n\n";
echo $script;
}
I am sharing this hoping this will give u some idea. Its obvious you have to : create infotips table, a model for that, and create a widget etc to fetch infotips related to your form fields . As someone suggested, if you are using Bootstrap, you have better way to do that.

Check if an existing value is in a database

I was wondering how I would go about checking to see if a table contains a value in a certain column.
I need to check if the column 'e-mail' contains an e-mail someone is trying to register with, and if something exists, do nothing, however, if nothing exists, insert the data into the database.
All I need to do is check if the e-mail column contains the value the user is registering with.
I'm using the RedBeanPHP ORM, I can do this without using it but I need to use that for program guidelines.
I've tried finding them but if they don't exist it returns an error within the redbean PHP file. Here's the error:Fatal error: Call to a member function find() on a non-object in /home/aeterna/www/user/rb.php on line 2433
Here's the code that I'm using when trying this:
function searchDatabase($email) {
return R::findOne('users', 'email LIKE "' . $email . '"');
}
My approach on the function would be
function searchDatabase($email) {
$data = array('email' => $email);
$user = R::findOne('users', 'email LIKE :email, $data);
if (!empty($user)) {
// do stuff here
} // end if
} // end function
It's a bit more clean and in your function
Seems like you are not connected to a database.
Have you done R::setup() before R::find()?
RedBeanPHP raises this error if it can't find the R::$redbean instance, the facade static functions just route calls to the $redbean object (to hide all object oriented fuzzyness for people who dont like that sort of thing).
However you need to bootstrap the facade using R::setup(). Normally you can start using RB with just two lines:
require('rb.php'); //cant make this any simpler :(
R::setup(); //this could be done in rb.php but people would not like that ;)
//and then go...
R::find( ... );
I recommend to check whether the $redbean object is available or whether for some reason the code flow has skipped the R::setup() boostrap method.
Edited to account for your updated question:
According to the error message, the error is happening inside the function find() in rb.php on line 2433. I'm guessing that rb.php is the RedBean package.
Make sure you've included rb.php in your script and set up your database, according to the instructions in the RedBean Manual.
As a starting point, look at what it's trying to do on line 2433 in rb.php. It appears to be calling a method on an invalid object. Figure out where that object is being created and why it's invalid. Maybe the find function was supplied with bad parameters.
Feel free to update your question by pasting the entirety of the find() function in rb.php and please indicate which line is 2433. If the function is too lengthy, you can paste it on a site like pastebin.com and link to it from here.
Your error sounds like you haven't done R::setup() yet.
My approach to performing the check you want would be something like this:
$count = count(R::find('users', 'email LIKE :email', array(':email' => $email)));
if($count === 0)
{
$user = R::dispense('users');
$user->name = $name;
$user->email = $email;
$user->dob = $dob;
R::store($user);
}
I don't know if it is this basic or not, but with SQL (using PHP for variables), a query could look like
$lookup = 'customerID';
$result = mysql_fetch_array(mysql_query("SELECT columnName IN tableName WHERE id='".$lookup."' LIMIT 1"));
$exists = is_null($result['columnName'])?false:true;
If you're just trying to find a single value in a database, you should always limit your result to 1, that way, if it is found in the first record, your query will stop.
Hope this helps

Does CDbcommand method queryAll() in yii return indexed entries only?

I am trying to retrieve data from a simple mySql table tbl_u_type which has just two columns, 'tid' and 'type'.
I want to use a direct SQL query instead of the Model logic. I used:
$command = Yii::app()->db->createCommand();
$userArray = $command->select('type')->from('tbl_u_type')->queryAll();
return $userArray;
But in the dropdown list it automatically shows an index number along with the required entry. Is there any way I can avoid the index number?
To make an array of data usable in a dropdown, use the CHtml::listData() method. If I understand the question right, this should get you going. Something like this:
$command = Yii::app()->db->createCommand();
$userArray = $command->select('tid, type')->from('tbl_u_type')->queryAll();
echo CHtml::dropdownlist('my_dropdown','',CHtml::listData($userArray,'tid','type'));
You can also do this with the Model if you have one set up for the tbl_u_type table:
$users = UType::model()->findall();
echo CHtml::dropdownlist('my_dropdown','',CHtml::listData($users ,'tid','type'));
I hope that gets you on the right track. I didn't test my code here, as usual, so watch out for that. ;) Good luck!