Question

PHP

$table = $_REQUEST['table'];

$query  = "SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.Columns where TABLE_NAME = $table";
$result = mysql_query($query);
$i = 0;
$arr = array();
while($row = mysql_fetch_array($result, MYSQL_NUM)) {
    $arr[$i] = $row['COLUMN_NAME'];
    $i++;
}
echo json_encode($arr);

Javascript

$("#verticalSelect").change(function() {
    var table = $("#verticalSelect option:selected").attr('value');

    $.post(PROCESSORFILE, {"task": "getTableDupeFields", "table": table}, function(data) {
        alert(data);
    }, "json");
});

I want take the array $arr from php and pass it to javascript where I can print its contents one by one. So far the code I have and variations of have returned errors, resource Id’s and often just nothing at all.

How can this be done in the most trite way?

Answer 1

Firstly, what’s wrong:

while($row = mysql_fetch_array($result, MYSQL_NUM)) {
    $arr[$i] = $row['COLUMN_NAME'];
    $i++;
}

You’re specifying MYSQL_NUM, which instructs mysql_fetch_array() to return a numeric array. With that flag set, you’d access the first value as $row[0]. To get an associative array and access the fields by name, you can either use mysql_fetch_assoc() instead, not pass any flag to mysql_fetch_array(), or pass it MYSQL_ASSOC.

Also, the $i variable you’re using is unnecessary – you can append elements to an array without specifying an index, like so:

 $arr[] = $row['COLUMN_NAME'];

Finally, your script is open to SQL injection when you insert a $_REQUEST variable directly into your sql. At the very least, you should run mysql_real_escape_string() on the value before adding it to the SQL.

$table = mysql_real_escape_string($_REQUEST['table']);