What is the correct syntax for this query?
var l=db.Fetch<article>("SELECT * FROM articles WHERE title LIKE '%#0%'", 'something');
Or should I use CHARINDEX?
May be
var l=db.Fetch<article>("SELECT * FROM articles WHERE title LIKE #0", "%something%");
I haven't tried this, but I think it is worth trying:
var l=db.Fetch<article>("SELECT * FROM articles WHERE title LIKE #0", "%" + "something" + "%");
If you have done your mappings (wich the T4 will do for you) then you could infact do it like so:
var l=db.Fetch<article>("WHERE title LIKE #0", "%something%");
Saves some typing :)
Can try like this also
var l=db.Fetch<article>("WHERE title LIKE #0", "%" + "something" + "%");
No need to write Select * From article, when you want to get all the fields. You are better off to use string.format to include the wildcards
var l=db.Fetch<article>("WHERE title LIKE #0", string.Format("%{0}%", yourVariable));
Articulo articulo = new Articulo();
articulo = db.SingleOrDefault<Articulo>("SELECT TOP (1) * FROM [Articulos] WHERE [CodigoEmpresa] = #0 and [CodigoArticulo] LIKE #1 ", CodigoEmpresa, codigoArticulo + "%");
Related
Hi I have the following statement that I execute using node-oracle
await connection.execute(`SELECT * FROM TABLE WHERE NAME LIKE '%And%'`)
But now I want to bind a parameter instead of using a hard coded value
const queryText = 'And';
await connection.execute(`SELECT * FROM TABLE WHERE NAME LIKE '%:queryText%'`, {queryText});
it throws Error: ORA-01036: illegal variable name/number
What is the correct way of binding a parameter here, since the documentation doesn't cover this situation?
Try with the following:
const queryText = 'And';
await connection.execute(
"SELECT * FROM TABLE WHERE NAME LIKE :queryText",
{
queryText: { dir: oracledb.BIND_IN, val: '%'+ queryText +'%', type: oracledb.STRING }
});
Use string concatenation:
SELECT * FROM TABLE WHERE NAME LIKE '%' || :queryText || '%'
Here is a working example.
let queryText = "John"
let sql = "SELECT * FROM TABLE WHERE NAME LIKE :queryText"
let binds = {queryTarget: {dir: oracledb.BIND_IN, val: queryText, type: oracledb.STRING}}
let result = await connection.execute(sql, binds, options)
Do not add '%' like the other people suggested.
In my controller
$agent = University::model()->findByPK($university_id);
I hope it will return value of a row of value.
I want a single attribute(field3) value say university_name, (with out using findByPK), how to get it
SELECT field3 FROM table [WHERE Clause]
Try this
$usercriteria = new CDbCriteria();
$usercriteria->select = "university_name";
$usercriteria->condition = "university_id=$university_id";
$university = University::model()->findAll($usercriteria);
echo $university->university_name;
Or simply do like u did first
$agent = University::model()->findByPK($university_id);
echo $agent-> university_name;
$agent = University::model()->findByPK($university_id);
echo $agent->university_name;
It should be like this:
$agent = University::model()->findByPK($university_id)->university_name;
Try this:
$university_name = University::model()->findByPK($university_id, array('select'=>'univeersity_name'))->university_name;
#Query: SElECT university_name FROM table_name where id=x;
Instead of
$university_name = University::model()->findByPK($university_id)->university_name;
#Query: SElECT * FROM table_name where id=x;
Second query returns all the fields. So better to avoid those fields are not necessary.
in case if you need to view on the yii _views, i have implemented this on my project
i put this inside my '_view.php' at /protected/views/myTable/
$agent = University::model()->findByPK($data->id_university/*this is the PK field name*/);
echo $agent->university_name /*university field name*/;
sorry again for my bad english :o
It can be done like this:
$agent = University::model()->findAllByAttributes(array('field3'),"WHERE `id` = :id", array(':id' => $university_id));
First argument of findByAttributes is an array of attributes you wish it to return. If left empty it returns all (*).
Whenever I tried to execute this sql query in a function module in drupal I am not able to get the results but when I try to execute this in MySQL I can view the result. My code looks like this :
function _get_subject_sub_category() {
$options = array();
$sql = "SELECT father.Subject_Code, child.Subject_Category
FROM {subjects} as child
INNER JOIN {subjects} as father ON (child.Parent_Category = father.Subject_Code
AND child.Level =2 )";
$result = db_query($sql);
foreach ($result as $row) {
$options[$row->father.Subject_Code] = $row->child.Subject_Category;
}
return $options;
}
The error I encountered is in the line $options[$row->father.Subject_Code] = $row->child.Subject_Category;`.
Any help will be highly appreciated.
Try changing this line:
$options[$row->father.Subject_Code] = $row->child.Subject_Category;
To this:
$options[$row->Subject_Code] = $row->Subject_Category;
The name of the table is not in the result. If you need to avoid confusion, you can use alias in your SQL query.
I don't know drupal and don't use php, but I would say :
remove
father. in $row->father.Subject_Code
and
child. in $row->child.Subject_Category
as father and child are just db aliases.
In SQL I can write
SELECT blah FROM Clients
Where #p1 Like '%'+lastname+'%'
How do I represent this with CreateCriteria in Nhibernate?
I've tried s.CreateCriteria<Client>.Add(Restrictions.Where<Client>(c => "something".Contains(c.LastName))
but get an error
System.Exception: Unrecognised method call: System.String:Boolean Contains(System.String)\r\n at NHibernate.Impl.ExpressionProcessor.ProcessCustomMethodCall(MethodCallExpression methodCallExpression)
I've also tried
s.CreateCriteria<Client>.Add(Restrictions.Where<Client>(c => "something".IndexOf(c.LastName) != -1))
but get
"variable 'c' of type 'TrinityFinance.Data.Entities.Client' referenced from scope '', but it is not defined"
Note the order is important here.
#p1 Like '%'+lastname+'%'
is not the same as
lastname Like '%'+#p1+'%'
s.CreateCriteria<Client>().Add(
Restrictions.InsensitiveLike( "LastName", "something", MatchMode.Anywhere))
Thanks to a friend I've solved my issue.
var searchCriteria = GetSession().CreateCriteria<Client>();
searchCriteria.Add(Expression.Sql(string.Format("'{0}' like '%' + {1} + '%'", p.ClientInputText,p.DbField)));
var results = searchCriteria.List<Client>();
For Case-Insensitive %Like% Search
Criteria criteria = session.createCriteria(Any.class);
criteria.add(Restrictions.ilike(propertyName, value, MatchMode.ANYWHERE);
criteria.list();
How to write SQL LIKE Query in drupal ,
SELECT title FROM { node } WHERE type='%s'
i want to add the LIKE CONDITION IN THAT
SELECT title FROM { node } WHERE type='%s' AND LIKE '%S%'
i think i writtern wrong like query formnat, can rewrite and tell me,
Just use % to escape.
$result = db_query('SELECT title FROM {node} WHERE type = "%s" AND title LIKE "%%%s%%"', 'type', 'title');
while ($row = db_fetch_object($result)) {
// do stuff with the data
}
Node type does not need escaping.
And here is an example with how to use LIKE in a dynamic query (Drupal 7 Only):
$query = db_select('node', 'n')
->fields('n', array('title'))
->condition('type', 'my_type')
->condition('title', '%' . db_like(search_string) . '%', 'LIKE');
$result = $query->execute()->fetchCol();
db_like() is used to escapes characters that work as wildcard characters in a LIKE pattern.
drupal_query replace %% to % and %s to value string
so your code will be
$sql = "SELECT title FROM node WHERE type='%%%s' AND title LIKE '%%%S%%'";
$type = "type to use in query";
$title = "title to use in query";
$result = db_result(db_query($sql, $type, $title));
OK, so you want the LIKE operator to refer to the title column. Use this query:
$sql = "SELECT title FROM node WHERE type='%s' AND title LIKE '%S%'";
$type = "type to use in query";
$title = "title to use in query";
$result = db_result(db_query($sql, $type, $title));
This is because the LIKE operator requires a column name to be specified. Otherwise, your database doesn't have any idea what value you want to perform the comparison on. See here.