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Editorial Team
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Asked: 2026-05-27T10:42:24+00:00

The goal: Connect to database using php to extract results Loop through results Store

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The goal:

  1. Connect to database using php to extract results
  2. Loop through results
  3. Store results in a way such that the are accessible during run-time

How I tried to do this:

  1. SQL query to get data
  2. Create function with inputs of column data that outputs this data as a javascript object
  3. Store javascript object in javascript array (not yet done)

For some reason (which may or may not be obvious to a more trained eye) I am running into a bug.

Note: database connection works perfectly, and the proper results are fetched with no issues. The issue is 100% the way I am trying to execute my javascript via php. I am assuming the php is seeing the var etc and is compiling it.

Please advise on the code below and if more specifics are needed I will happily expand:

        <?
        if (!isset($_COOKIE['authPw'])){
        echo '<script>window.location="http://www.google.com"</script>';
        }
        ?>
        <html>
            <head>
            </head>

            <body>

            <?

            function jsEcho($i,$cN,$fN,$lN,$pN,$eM){
                $jString="var appObj = new Object();";
                $jString+="appObj.id=" + $i + ";";
                $jString+="appObj.companyName=\'" + $cN + "\';";
                $jString+="appObj.firstName=\'" + $fN + "\';";
                $jString+="appObj.lastName=\'" + $lN + "\';";
                $jString+="appObj.phoneNumber=\'" + $pN + "\';";
                $jString+="appObj.eMail=\'" + $eM + "\';";  
                //echo "1ONE1".$jString."2TWO2";
                return $jString;
            }

            $host="localhost"; // Host name 
            $username="abc"; // Mysql username 
            $password="def"; // Mysql password 
            $db_name="ghi"; // Database name 
            $tbl_name="jkl"; // Table name





            $link=mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
            mysql_select_db("$db_name")or die("cannot select DB");

            $mySQL="select * from ".$tbl_name.";";
            echo $mySQL;
            $result=mysql_query($mySQL);
            $jS="var aApps=new Array();";

                while ($row = mysql_fetch_array($result))
                {

                    $iD=$row["id"];
                    $companyName=$row["companyName"];
                    $firstName=$row["firstName"];
                    $lastName=$row["lastName"];
                    $phone=$row["phone"];
                    $email=$row["email"];

                    $jS+=jsEcho($iD,$companyName,$firstName,$lastName,$phone,$email);
                         }

            ?>


            </body>
            <?
            echo '<script>';
                    echo $jS;
            echo '</script>';
            ?>
        </html>
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1 Answer

  1. Editorial Team

    Your script is after the body element closes. That’s not valid html. Possibly it is not being evaluated. I suspect the output is also not valid javascript either.

    Also, your jsEcho method is verbose, unsafe, and unnecessary. Use this pattern:

    <?php
// prepare all your data as a single PHP object or array
$jsdata = array();
while ($row = mysql_fetch_object($result)) {
    $jsdata[] = $row;
}

// now encode to JSON
$jsenc = json_encode($jsdata);

// now html-escape it
?>
<script>
var allMyData = <?php echo htmlspecialchars($jsenc, ENT_NOQUOTES, 'utf-8')?>;
</script>
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