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Home/ Questions/Q 6060955
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T08:53:01+00:00 2026-05-23T08:53:01+00:00

I have a situation where I would like to use a php file query.php

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I have a situation where I would like to use a php file “query.php” to look take either a $_POST or $_GET value as the MySQL query. It looks like this:

<?php
//verify data has been sent via POST or GET and set strings
//find supplied table name
if(isset($_POST['tblName'])){
    $strSuppliedTableName = $_POST['tblName'];
}
if(isset($_GET['tblName'])){
    $strSuppliedTableName = $_GET['tblName'];
}
else{
    $strSuppliedTableName = 'roles';
}

//find supplied field name or default to all fields in the table
if(isset($_POST['fieldName'])){
    $strSuppliedFieldName = $_POST['fieldName'];
}
else if(isset($_GET['fieldName'])){
    $strSuppliedFieldName = $_GET['fieldName'];
}
else{
    $strSuppliedFieldName = '*';
}

//query db
$query = 'SELECT ' . $strSuppliedFieldName . ' FROM ' . $strSuppliedTableName;

$results = mysql_query($query) or die(mysql_error());
?>

Following that, I want to include this file “query.php” in another file that will manage the results. I’m trying to make this as modular as possible.

<?php
require_once("query.php?tblName=classes");
......... (while loop, yadi yadi

However, I recieve an error:

Warning: require_once(query.php?tblName=classes) [function.require-once]: failed to open stream: No such file or directory

Is it not acceptable to pass GET values to your included file? PHP won’t process this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T08:53:02+00:00Added an answer on May 23, 2026 at 8:53 am

    You do not need to pass variables in a get or POST like way when including or requiring files, the variables are shared between the files, as long as the values are set before the including happens.

    i.e.:

    file1.php called as file1.php?var2=value2

    <?php
    $var1 = "value1";
    $var2 = $_GET['value'];
    include "file2.php";
    ?>
    

    file2.php:

    <?php
    echo $var1.' '.$var2;
    ?>
    

    will output:

    value1
    value2

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