CODEWITHSUNDEEP
phpmysql
Code Grasshopper
Code Grasshopper

Reputation: 620

Correct syntax to update a table in a database using PHP

What would the correct syntax to update a table be while using PHP to make the connection? If I manually change a row on Wamp server it shows me this:

UPDATE  `proyecto`.`labs` SET  `estado` =  '2' WHERE  `labs`.`idlab` =1;

But a quick Ctrl+C and Ctrl+V on my code doesn't do anything, nor does it trigger any error. For reference here is my PHP code:

$db_hostname = 'localhost';
$db_database = 'XXX';
$db_username = 'XXX';
$db_password = 'XXX';

$db_server = db_init();

function db_init(){
 global $db_hostname,
 $db_database,
 $db_username,
 $db_password;

$db_server = mysql_connect($db_hostname, $db_username, $db_password);

if(!$db_server)
die ("Unable to connect with MySQL: " . mysql_error());

 mysql_select_db($db_database)
 or die ("Unable to connect with database " . mysql_error());

return $db_server;

}

function do_query($query){
global $db_server;

$result = mysql_query($query , $db_server);

if( !$result)
    die("Failed sentence: " . mysql_error());

    return $result;
}

This is the function I am trying to invoke, it is being call via jQuery, just in case I am adding the code below:

function labStatusEliminar(){

if(isset($_POST['pId'])){

    $idEliminar = $_POST['pId'];

    $query = "UPDATE bk.estado FROM labs as bk  WHERE  bk.idlab = $idEliminar";
 }
}

jQuery:

$(".btnEliminar").click(function() {
  var idAttrE = $(this).attr('id');
  var idEliminar = parseInt(idAttrE.substring(12));

 var request = $.ajax({
  url: "includes/functionsLabs.php",
  type: "post",
  data: {

    'call': 'labStatusEliminar',
    'pId':idEliminar},

    dataType: 'html',

    success: function(response){
    $('#info').html(response);
    }
  });
});

Upvotes: 0

Views: 83

Answers (2)

AyB
AyB

Reputation: 11665

The correct syntax for UPDATE is:

UPDATE `table_name`
SET `column_name` = 'value', `column_name`='value'...
WHERE `column_name` = 'value'

Upvotes: 1

vhu
vhu

Reputation: 12798

Looks like your UPDATE statement is missing the SET part.

 $query = "UPDATE bk.estado FROM labs as bk  WHERE  bk.idlab = $idEliminar";

That probably should be 

 $query = "UPDATE labs SET estado=2 WHERE labs.idlab = $idEliminar";

I'd also suggest you look into mysqli or PDO and use prepared statements instead. You code is currently vulnerable to SQL injection.

Upvotes: 1

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