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Select where clause with "IN"

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Join Date: Oct 2006

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Reputation: assgar is an unknown quantity at this point

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assgar

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Select where clause with "IN"

May 25th, 2008

Hi

I would like to put a list of ids in an array
and loop to select data from a table using
"IN" in the where clause.

I am not having any success getting this to work.
Any suggestions?

Note:I am using Mysql database, apache, linux/Windows

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
[php]
<?
//array with list of friends to display
$friend = array('1','2','3','4');
 
$query = "SELECT first, last, number
	  FROM friend
	  WHERE id IN '$friend'";
$result = mysqli_query($mysqli, $query);
[php]
<?
//array with list of friends to display
$friend = array('1','2','3','4');

$query = "SELECT first, last, number
	  FROM friend
	  WHERE id IN '$friend'";
$result = mysqli_query($mysqli, $query);

[/php]

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ShawnCplus ShawnCplus is offline

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Re: Select where clause with "IN"

May 26th, 2008

I'm pretty sure it has to be surrounded with quotes, ie.,

 Help with Code Tags 
 sql Syntax (Toggle Plain Text) 
SELECT first,last, number FROM friend WHERE id IN ('1','2','3','4')
SELECT first,last, number from friend WHERE id IN ('1','2','3','4')

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nav33n

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Senior Poster

Re: Select where clause with "IN"

May 26th, 2008

•

Originally Posted by assgar

Hi

I would like to put a list of ids in an array
and loop to select data from a table using
"IN" in the where clause.

I am not having any success getting this to work.
Any suggestions?

Note:I am using Mysql database, apache, linux/Windows
Help with Code Tags

PHP Syntax (Toggle Plain Text)
[php]
<?
//array with list of friends to display
$friend = array('1','2','3','4');
 
$query = "SELECT first, last, number
	  FROM friend
	  WHERE id IN '$friend'";
$result = mysqli_query($mysqli, $query);
[php] <? //array with list of friends to display $friend = array('1','2','3','4'); $query = "SELECT first, last, number FROM friend WHERE id IN '$friend'"; $result = mysqli_query($mysqli, $query);
[/php]

You can't do it this way because $friend is an array. So, when you execute the query, it will be,

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$query = "SELECT first, last, number
	  FROM friend
	  WHERE id IN 'array'";
$query = "SELECT first, last, number
	  FROM friend
	  WHERE id IN 'array'";

which is wrong. You can use implode function to join the array.
$friend_implode = implode(",",$friend); Then use it in your query.

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$query = "SELECT first, last, number
	  FROM friend
	  WHERE id IN (".$friend_implode.")";
$query = "SELECT first, last, number
	  FROM friend
	  WHERE id IN (".$friend_implode.")";