Reputation: 135
I am trying to use a check box as input in form where the records are printed from a mysql table. Following is the code which I have used :-
<body>
<form name='checkbox_text' method="post" action="submit_checkbox.php">
<select name="candidate_name" id="option2">
<b><h3>Select the candidates :-</h3></b>
<label class="q" for="qq1" </label>
<input type="hidden" name="qq1[]" value="null">
<?php
mysql_connect("localhost", "root", "root1") or die (mysql_error ());
mysql_select_db("my_database") or die(mysql_error());
$strSQL = "SELECT Name FROM Candidate ORDER BY Name";
$rs = mysql_query($strSQL);
while($row = mysql_fetch_array($rs))
{
$man = $row['Name'];
Print ("<option>");
echo $man;
Print ("</option>");
}
mysql_close();
?>
</select>
</form>
</body>
I am able to print all the table contents but I need a checkbox in front of each name, user shall select few names and pass those values in submit_checkbox.php
Upvotes: 1
Views: 203
Reputation: 425
maybe something like this?
while($row = mysql_fetch_array($rs)) {
$man = $row['Name'];
echo '<input type="checkbox" value="'.$man.'" name="checkbox[]" />';
echo $man;
}
if you want multiple values after you submit form, you have to use [] in name attribute. you'll get an array with all selected names after you post the form.
Upvotes: 1
Reputation: 31739
Try with Multiple
select
<h3>Select the candidates :-</h3>
<select name="candidate_name[]" id="option2" multiple>
<?php
mysql_connect("localhost", "root", "root1") or die (mysql_error ());
mysql_select_db("my_database") or die(mysql_error());
$strSQL = "SELECT Name FROM Candidate ORDER BY Name";
$rs = mysql_query($strSQL);
while($row = mysql_fetch_array($rs))
{
$man = $row['Name'];
Print ("<option>");
echo $man;
Print ("</option>");
}
mysql_close();
?>
</select>
Upvotes: 1