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Home/ Questions/Q 7248031
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T22:05:04+00:00 2026-05-28T22:05:04+00:00

I wonder if anyone can help – I want to select a table and

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I wonder if anyone can help – I want to select a table and create Javascript a array with the return values. Here’s what I have so far:

con = mysql_connect("localhost","usrname","password");

if (!$con) {
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("my_db", $con);
$ql = "SELECT theMessage FROM email_message";
$result = mysql_query($ql) or die(mysql_error());

while ($row = mysql_fetch_array($result)) {
    $allMessage = $row['theMessage'] . " ";
}

$arr = array($allMessages);
$script = '<script>var newArr = new Array(' . implode(' ', $arr) . ');</script>';
echo $script;

But instead, it just shows me empty array like this: var newArr = new Array();

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  1. Editorial Team
    Editorial Team
    2026-05-28T22:05:05+00:00Added an answer on May 28, 2026 at 10:05 pm

    Change the line

    $allMessage = $row['theMessage'] . " ";
    

    to

    $allMessages[] = $row['theMessage'] . " ";
    

    (adding a s to the variable name, and adding the [])
    because you are overwriting your result every time you read a new line, and afterwards read from a different (empty) variable!

    Now you should implode the array not with spaces, you should implode it with “, ” or with “‘, ‘”, depending on what data is stored.

    You could also put the output directly into the fetch-loop, but thats just an idea for you.

    What you really should consider about is reading the json_encode() and json_decode():

    http://www.php.net/manual/en/function.json-encode.php

    http://www.php.net/manual/en/function.json-decode.php

    Hope i could help.

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