CODEWITHSUNDEEP
phpmysqlstored-proceduresxampp
Dobro
Dobro

Reputation: 41

Calling MySQL Stored Procedure through PHP

I've installed XAMPP and have been attempting to interrogate my database using PHP although I keep getting the same error. 

<?php
$servername = "localhost";
$username = "root";
$password = "secret";

// Create connection
$conn = new mysqli($servername, $username, $password);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

// Perform queries 
$result = mysqli_query($conn, 
    "use edgeserver; call ShowAll") or die("Query fail: " . mysqli_error());

  //loop the result set
while ($row = mysqli_fetch_array($result)){   
     echo $row[0] . " - " . + $row[1]; 
  }

$conn->close();

?>

However this presents me with the following error:

Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\xampp\htdocs\amit\ShowAll.php on line 16
Query fail:

Upvotes: 0

Views: 826

Answers (3)

Dobro
Dobro

Reputation: 41

Thanks, it ended up being a combination of what you both said:

<?php   

// Create connection
$conn = new mysqli('localhost', 'root', 'secret', 'edgeserver');

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

// Perform queries 
$result = mysqli_query($conn, 
    "call ShowAll") or die("Query fail: " . mysqli_error($conn));

  //loop the result set
while ($row = mysqli_fetch_array($result)){   
     echo $row[0] . " - " . + $row[1]; 
  }

$conn->close();

Great response, hopefully this will be the base to me developing some good code.

Upvotes: 1

Stefan C
Stefan C

Reputation: 123

I think the problem is from use edgeserver; call ShowAll.

use edgeserver, where edgeserver is the database that you want to use. call ShowAll - what you try to accomplish with this?

Your new mysqli must be like this:

$mysqli = new mysqli('localhost', 'my_user', 'my_password', 'my_db');

source: http://php.net/manual/ro/mysqli.construct.php

And on mysqli_query you must do the queries like "Select * from table".

Upvotes: 0

chandresh_cool
chandresh_cool

Reputation: 11830

Mysql_error requires db connection to be passed. Try below:

mysqli_error($conn);

Upvotes: 1

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