PHP variable variables in MySQL query

Question

This code works fine: $batsman1name is inserted as a new value in the correct row in the database.

for($count = 1; $count <= 22; ++$count)
{
    $setbattingid = 'batsman' . $count . 'battingid';
    $$setbattingid = mysql_real_escape_string($_POST[$setbattingid]);
    $setname = "batsman" . $count . "name";
    $$setname = mysql_real_escape_string($_POST[$setname]);
    $query = "UPDATE batting_new SET batsmanname = NULLIF('$batsman1name', '') WHERE battingid = '$batsman1battingid'";
    $result = mysql_query($query);
    if (!$result) die ("Database access failed: " . mysql_error());
}

With this code the database is not updated, but the varaible variables $$setname and $$setbattingid do contain the same values as $batsman1name and $batsman1battingid above.

for($count = 1; $count <= 22; ++$count)
{
    $setbattingid = 'batsman' . $count . 'battingid';
    $$setbattingid = mysql_real_escape_string($_POST[$setbattingid]);
    $setname = "batsman" . $count . "name";
    $$setname = mysql_real_escape_string($_POST[$setname]);
    $query = "UPDATE batting_new SET batsmanname = NULLIF('$$setname', '') WHERE battingid = '$$setbattingid'";
    $result = mysql_query($query);
    if (!$result) die ("Database access failed: " . mysql_error());
}

Any ideas? Let me know if I haven't explained my question very well? Thanks.

soju · Accepted Answer · 2012-10-09 08:59:07Z

1

You should use :

$query = "UPDATE batting_new SET batsmanname = NULLIF('${$setname}', '') WHERE battingid = '${$setbattingid}'";

As described here : http://php.net/manual/language.variables.variable.php

answered Oct 9, 2012 at 8:59

soju

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