PHP mysql connection

[Louffeman]

Hi,I have a problem of my shopping cart script,my server indicate the error in this line: $produit = $result->fetch_assoc();error: Fatal error: Call to a member function fetch_assoc() on a non-object in C:wampserver32wwwsaleajout_panier.php on line 7je d'ont know how to solve it,anyone can help me please ?<?phpsession_start();$mysql = new mysqli('localhost', 'root', '', 'testebase');$result = $mysql->query('SELECT * FROM trybase WHERE name= '.$_POST['name']);$produit = $result->fetch_assoc();require_once 'panier.php';$valeur = array('name'=>$produit['name'],'price'=>$produit['price'],'quantity'=>$_POST['quantity']);$panier->set($_POST['name'], $valeur);header('location: votre_panier.php');?>

[JamesB]

you need apostrophes around the value to indicate a MySQL string:

 

 

$result = $mysql->query("SELECT * FROM trybase WHERE name= '".$_POST['name']."'");

[Louffeman]

@JamesBthank you for your reply, I did as you said,but now, the error is in the line 5: $mysql = new mysqli('localhost', 'root', '', 'testebase');Parse error: syntax error, unexpected T_STRING, expecting T_VARIABLE in C:wampserver32wwwsalepanier.php on line 5

[JamesB]

do you have an apostrophe in your password? if so you need to escape all apostrophes used by putting a backslash before them, eg.

 

$mysql = new mysqli('localhost', 'root', 'my'pass'word', 'testebase');

[Louffeman]

yes, as you see:$mysql = new mysqli('localhost', 'root', '', 'testebase');

[Ingolme]

Make sure that the syntax is correct on the lines before line 5. I don't see an error on that line in particular.

[Louffeman]

@Ingolmethe server has no any error indicate before the line 5

[JamesB]

yes, as you see:$mysql = new mysqli('localhost', 'root', '', 'testebase');

When i said to escape the apostrophes, this excludes the first and last per string, as shown in my example:

 

'my'pass'word'

 

 

 

@Ingolmethe server has no any error indicate before the line 5

Sometimes the error line is off by 1 line, can you post your previous line of code.

[Louffeman]

I have corrected my script ajout_panier.phpnow, it is OK, but the last script(votre_script) still has problem in the line 7:Undefined variable: listproduit in C:wampserver32wwwsalevotre_panier.php on line 7the line 7 is: <?php if(!$listproduit){?>the output is the cart is empty, which mean's it didn't receive anything ?it is votre_script.php<?phpsession_start();require_once 'panier.php';$panier = new Panier('produits');?><?php if(!$listproduit){?><p> The cart is empty </p><?php } else {?><table border="1" width="50%"><tr><td>Name</td><td>Price</td><td>Quantity</td></tr><?php foreach($listproduit as $produit) { ?><tr><td><?php print $produit['name'] ?></td><td><?php print $produit['price'] ?></td><td><?php print $produit['quantity'] ?></td></tr><?php } ?></table><?php } ?><p><a href="index.php">Home</a></p>

[JamesB]

You can either set $listproduit to false before it is conditionally set to true, or use if(!isset($listproduit))

[Louffeman]

I have already solved the problem, thanks a lot

[justsomeguy]

$result = $mysql->query("SELECT * FROM trybase WHERE name= '".$_POST['name']."'");

This is terrible practice, this is no better than using the deprecated mysql extension. Use prepared statements so that you don't have problems like this.https://www.google.com/search?client=opera&q=mysqli+prepared+statements+tutorial&sourceid=opera&ie=UTF-8&oe=UTF-8