How to create a table in database using jquery

Question

I have a button that when pressed will create a table in the database. I tried this

index.html

<button id="loadButton" type="button" class="btn btn-success" style="margin-top:5px;">Carregar Base de Dados</button>

dash.js

$('#loadButton').click(function() {
    $.ajax({
        url: 'connectdb.php'
    });
});

connectdb.php

<?php

$server = 'localhost';
$user = 'root';
$password = '*****';
$database = 'test';

$con = mysql_connect($server, $user, $password, $database);

if (!$con) {
  die('error: ' . mysql_error($con));
}

// Create table
$sql="CREATE TABLE Post( 
    id_post int Primary Key, 
    titulo varchar(100) not null, 
    imagem longblob,
    descricao varchar(1000) not null,
    hashtag varchar(100) not null
)";

// Execute query
mysql_query($con,$sql))
?>

?>

but when I check my database table was not created. I'm new in jquery, what I'm doing wrong. can someone help me?

mysql_query -> mysql_query($sql,$link), mysqli_query -> mysqli_query($link,$sql). Update to mysqli. Also, you may want to use CREATE TABLE IF NOT EXISTS Posts, as if you click again it will overwrite your table. — Sean
– Sean, Commented Aug 2, 2014 at 16:54
Why would you create a table for every post? Additionally, this is bad practice anyway. I would also recommend using English names for columns. While end user localized text is a good thing, IMHO code should always use English, as it's the language of business in today's world. — Zarathuztra
– Zarathuztra, Commented Aug 2, 2014 at 17:00

Max Hutschenreiter · Accepted Answer · 2014-08-02 17:04:01Z

1

Try this and if you want to be shure the php code is called replace all the code with an simple echo and check the response.

$server = 'localhost';
$user = 'root';
$password = '*****';
$database = 'test';

$db = new mysqli($server, $user, $password, $database);

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$db->query("
CREATE TABLE Post( 
    id_post int Primary Key, 
    titulo varchar(100) not null, 
    imagem longblob,
    descricao varchar(1000) not null,
    hashtag varchar(100) not null
)");

answered Aug 2, 2014 at 17:04

Max Hutschenreiter

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Comments

moskito-x · Accepted Answer · 2014-08-02 17:41:48Z

1

First : Please use MySQLi* mysql_* is deprecated.

You have errors in your .php code

mysql_query($con,$sql))

one ) of )) to much
you should have a ; at the end of your command
wrong order $con,$sql

this is OK

mysql_query($sql,$con);

sometimes you should use the database here 'test' in your query

$database = 'test';
...
$sql="CREATE TABLE test.Post(

you should test your result

$result = mysql_query($sql,$con);
if (!$result) {
    $message  = 'invalid query: ' . mysql_error() . "\n";
    $message .= 'your query: ' . $sql;
    echo $message;
    die($message);
}

To search for errors this is very helpful !

answered Aug 2, 2014 at 17:41

moskito-x

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Comments

Jake · Accepted Answer · 2014-08-02 17:00:57Z

-1

go to connectdb.php directly e.g. http://www.example.com/connectdb.php, fix all the errors and make sure it is working first.

answered Aug 2, 2014 at 17:00

Jake

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3 Comments

Jake Over a year ago

How would I know what errors he is getting when he did not post his errors messages? I have provided a debugging method as I am guessing he has errors in connectdb.php although I cannot know for sure.

Zarathuztra Over a year ago

Ask that in a comment. I don't consider this a valid answer.

Jake Over a year ago

I respect your judgement, but I don't intend to know what error he gets as I think he is capable of fixing the errors, just that he cannot see the error messages because it is an AJAX call and he is new to jQuery. Thus I am certain that by going directly to connectdb.php will answer "what I'm doing wrong. can someone help me?"

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