Mysqli insert not working properly

[cutielou22]

Not sure if this is the right place for this or not but . . . I am having trouble with a mysqli code - I am not sure what the problem is and it doesn't give me any errors. It simply does not insert the information to the database.

 

Here it is:

$addband = $mysqli->prepare("INSERT INTO bands (id, bandname, bandurl, bio, status, twitter, facebook, myspace, youtube, itunes, website, hometown, genre, genre1, genre2, genre3, genre4, recordlabel, email, manager, bookingagent, extra, tags) VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)") or die(mysqli_error());
$addband->bind_param('sssbssssssssssssssssss', $blank, $bandname, $bandurl, $bandbio, $status,  $twitter, $facebook, $myspace, $youtube, $itunes, $website, $hometown, $genre, $genre1, $genre2, $genre3, $genre4, $recordlabel, $email, $manager2, $bookagent2, $extra, $tags);
$addband->execute();
$addband->fetch();
$addband->close();

[peipst9lker]

Is your $mysqli-resource properly connected to the database?

 

If so, please post the output of

var_dump($addband, $mysqli);

(place it under your prepare statement)

 

 

Also It's useful to check for the return value prepare() gives out e.g.

 

if ($select = $mysqli->prepare("SELECT foo FROM bar")) {
    // bindparam,execute, etc.
} else {
    echo 'prepare() failed!';
}

[cutielou22]

It still didn't add data to the database and no errors came up.

 

Here is what I tried:

if ($addband = $mysqli->prepare("INSERT INTO band (id, bandname, bandurl, bio, status, twitter, facebook, myspace, youtube, itunes, website, hometown, genre, genre1, genre2, genre3, genre4, recordlabel, email, manager, bookingagent, extra, tags) VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)")) {
var_dump($addband, $mysqli);
$addband->bind_param('sssbssssssssssssssssss', $blank, $bandname, $bandurl, $bandbio, $status,  $twitter, $facebook, $myspace, $youtube, $itunes, $website, $hometown, $genre, $genre1, $genre2, $genre3, $genre4, $recordlabel, $email, $manager2, $bookagent2, $extra, $tags);
$addband->execute();
$addband->fetch();
$addband->close();

header("Location: addband.php?note=$bandname+has+been+added.");
} else {
    echo 'prepare() failed!';
}

[jazzman1]

Start from very beginning:

 

<?php
$mysqli = new mysqli("localhost", "user", "pass", "database");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

// continue with prepare statement

[cutielou22]

Nope. Nothing. :/

[jazzman1]

Check out this guide -> http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

[jazzman1]

Take a note:

 

Also, consider the use of the MySQL multi-INSERT SQL syntax for INSERTs. For the example, multi-INSERT requires less round-trips between the server and client than the prepared statement shown above

<?php
if (!$mysqli->query("INSERT INTO test(id) VALUES (1), (2), (3), (4)")) {
    echo "Multi-INSERT failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
?>

 

[jcbones]

bind_param will fail, as you only passed 22 types (1st param), but 23 variables to it.

You are also sending your fourth variable as a blob, which chunks it out.  Pretty sure you have to send that with mysqli-stmt.send_long_data(), but I could be wrong on that.

[cutielou22]

@jcbones You were right I checked that over and over again thinking that may have been a problem and still didn't catch it.

 

@jazzman1 Now it does give an error using:

 

if (!$mysqli->query("INSERT INTO band (id, bandname, bandurl, bio, status, twitter, facebook, myspace, youtube, itunes, website, hometown, genre, genre1, genre2, genre3, genre4, recordlabel, email, manager, bookingagent, extra, tags) VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)")) {
    echo "Multi-INSERT failed: (" . $mysqli->errno . ") " . $mysqli->error;
} else{

if ($addband = $mysqli->prepare("INSERT INTO band (id, bandname, bandurl, bio, status, twitter, facebook, myspace, youtube, itunes, website, hometown, genre, genre1, genre2, genre3, genre4, recordlabel, email, manager, bookingagent, extra, tags) VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)")) {
var_dump($addband, $mysqli);
$addband->bind_param('sssssssssssssssssssssss', $blank, $bandname, $bandurl, $bandbio, $status, $twitter, $facebook, $myspace, $youtube, $itunes, $website, $hometown, $genre, $genre1, $genre2, $genre3, $genre4, $recordlabel, $email, $manager2, $bookagent2, $extra, $tags);
$addband->execute();
$addband->fetch();
$addband->close();

header("Location: addband.php?note=$bandname+has+been+added.");
} else {
    echo 'prepare() failed!';
}
}

 

The error given:

Multi-INSERT failed: (1064) You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)' at line 1

[cutielou22]

I got it working! Not sure exactly how though . . . thanks for the help you guys!