PHP SQL Query Fails to Execute

Question

<?php
$db = new mysqli("localhost", "HIDDEN", "HIDDEN", "HIDDEN");
if ($db->connect_error) {
    die("Failed to connect.");
}
if (isset($_POST["title"]) && isset($_POST["description"]) && isset($_POST["url"])) {
    $title = $db->real_escape_string($_POST["title"]);
    $description = $db->real_escape_string($_POST["description"]);
    $url = $db->real_escape_string($_POST["url"]);
    $sql = "INSERT INTO video (name, description, submission_date)
    VALUES ('{$title}', '{$description}', CURDATE());
    INSERT INTO video_source (video_id, url)
    VALUES (LAST_INSERT_ID(), '{$url}');";
    if ($db->query($sql) === TRUE) {
        echo "Successfully added.";
    } else {
        echo "Query failed.<br><br>Data: {$title} {$description} {$url}";
    }
} else {
    echo "Data not set.";
}
$db->close();?>

Outputs "Query failed." with the data I entered. Replacing variables such as title with constants still has the same problem. I tried the query in PHPMyAdmin and it worked fine (with constants).

It seems to be unhappy with setting the value of video_id.

When localhost is properly enclosed in quotes the code still outputs the same error. — Ocordu
– Ocordu, Commented Jul 16, 2015 at 18:04

Jay Blanchard · Accepted Answer · 2015-07-16 18:14:49Z

2

Anytime you're running multiple queries with MySQLi you should use multi_query():

$db->multi_query($sql)

In addition, LAST_INSERT_ID() in your second query is not returning any sort of value. If you're looking for the last inserted value of the 1st query you have to return that prior to running the second query.

answered Jul 16, 2015 at 18:14

Jay Blanchard

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1 Comment

Ocordu Over a year ago

That worked, thank you :) Oddly enough when I just used the second query with only constants it still didn't work :/

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