Multiple dropdown values inserting in a single row not in multiple row
Question
In this code when i selected multiple values from dropdown it is stoing in a single row but i like to apply value in each row so please anybody help me. Actual result:
id game
1. cricket,football,tennis
Expecting result:
id game
1 cricket
2 football
code -
<html>
<body>
<?php
if(isset($_POST['submit']))
{
$query=mysql_connect('localhost','root','');
mysql_select_db("freeze",$query);
$choice=$_POST['game'];
$choice1=implode(',',$choice);
mysql_query("insert into tb values('','$choice1')");
}
?>
<form method="post" action="multipleselect.php">
Select your favourite game:<br/>
<select name="game[]" multiple="multiple">
<option>Football</option>
<option>Volleyball</option>
<option>Badminton</option>
<option>Cricket</option>
<option>Cricket1</option>
<option>Cricket2</option>
<option>Cricket3</option>
<option>Cricket34</option>
</select>
<input type="submit" name="submit">
</form>
</body>
</html>
4 answers
solution1
3 2013-08-10 05:13:09
First of all, please don't use mysql_* as it's deprecated, use mysqli_ or PDO instead.
now if you just want the values of options then do it like this
<select name="game[]" multiple="multiple">
<option value="1">Football</option>
<option value="2">Volleyball</option>
...
</select>
this way it'll give you 1,2,.... . Hope that's what you're looking for.
and if you're looking for query like this
INSERT INTO tb (`game`) VALUES ('Football'),('Volleyball')
assuming that id field is auto-incremented, then change the code as follows:
html code
<select name="game[]" multiple="multiple">
<option>Football</option>
<option>Volleyball</option>
...
</select>
php code
$choice=$_POST['game'];
$sql = "INSERT INTO tb (`game`) VALUES ";
$sqlValues= null;
foreach($choice as $ch) {
$sqlValues .= "('$ch')," ;
}
$sql.=rtrim($sqlValues, ",");
echo $sql;
this way you could get
id game
1 cricket
2 football
solution2
1 ACCPTED 2013-08-10 05:14:02
It is inserting them all in 1 row, because you are imploding your $_POST['game']
$choice=$_POST['game'];
$choice1=implode(',',$choice);
mysql_query("insert into tb values('','$choice1')");
You need to loop over each $_POST['game'] value
foreach($_POST['game'] as $choice){
mysql_query("insert into tb values('','$choice')");
}
note, your query is open to SQL injection, make sure to sanitize your data before inserting into your db. Also, mysql_* are deprecated, so you should update to mysqli or PDO - http://php.net/manual/en/mysqlinfo.api.choosing.php
solution3
0 2013-08-10 05:11:22
Are there any error messages being displayed? Insert the following code just before the mysql_query statement:
echo "insert into tb values('','$choice1')";
what is being displayed?
solution4
0 2013-08-10 05:17:32
<?php
if(isset($_POST['submit']))
{
$query=mysql_connect('localhost','root','');
mysql_select_db("freeze",$query);
$choice=$_POST['game'];
$choice1=implode(',',$choice);
foreach($choice1 as $choice)
{
mysql_query("insert into tb values('','$choice')");
}
}
?>
You just try this code i hope it works for you. If not then please reply me.
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