Error with this php code

Question

I am seeing this error on my website.

Deprecated: mysql_query(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in /home/u857740002/public_html/datafetch.php on line 3

Deprecated: mysql_query(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in /home/u857740002/public_html/datafetch.php on line 23

I am using this code.

Can anyone correct these two lines.

$siteSettings = mysql_query("select * from pu_settings where sid = '1' ");
$siteSoft = mysql_query("select * from pu_softwares where slug = '$slug' ");

possible duplicate of [The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead](http://stackoverflow.com/questions/13944956/the-mysql-extension-is-deprecated-and-will-be-removed-in-the-future-use-mysqli) — Stephen Kennedy, Jul 03 '15 at 07:28
This is a warning you can still use mysql_query but its old and unsafe, use http://php.net/manual/en/mysqli.query.php instead, or the next generation http://php.net/manual/en/class.pdo.php. — Ogelami, Jul 03 '15 at 07:28

bnahin · Answer 1 · 2015-07-03T08:37:47.297

3

MySQL_* is deprecated and is going to be removed very soon with PHP. Use MySQLi_*. See: MySQLi

Use this instead:

$link = mysqli_connect([dbhost],[dbusername],[dbpass],[dbname]);

$siteSettings = mysqli_query($link,"select * from pu_settings where sid = '1' ");
$siteSoft = mysqli_query($link,"select * from pu_softwares where slug = '$slug' ");

edited Jul 03 '15 at 08:37

answered Jul 03 '15 at 07:29

bnahin

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Is it right what you wrote because my knowledge says that mysqli_query need two parameters that is connection object and query. You are giving query only, where is connection object? – Alive to die - Anant Jul 03 '15 at 07:36
it solved my problem. but now a blank screen is showing up. how to solve this problem – Zeeshan Jul 03 '15 at 08:05
mysqli_query() takes two parameters. how it solved. if you run this codeit will give you error. – Alive to die - Anant Jul 03 '15 at 08:23
Forgot the first parameter.The first parameter should be the link to the connection, returned from mysqli_connect(). `mysqli_query($link,$query)`. View the edited answer. – bnahin Jul 03 '15 at 08:35
you are right @anantkumarsingh . my site is showing blank screen only. how to solve it? – Zeeshan Jul 03 '15 at 10:55
Try to use this on each page at starting after ` – Alive to die - Anant Jul 03 '15 at 11:16

Alive to die - Anant · Answer 2 · 2017-03-02T04:29:59.980

mysql_* is deprecated+removed library so use mysqli_* or PDO:-

suppose your database connection object is $conn

$siteSettings = mysqli_query($conn,"select * from pu_settings where sid = '1' ")or die(mysqli_error($conn));
$siteSoft = mysql_query($conn,"select * from pu_softwares where slug = '$slug' ")or die(mysqli_error($conn));

The code to create a connection using mysqli_* is:-

$conn=mysqli_connect("hostname","username","password","databasename");

if (mysqli_connect_errno()){
 echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

score 1 · Answer 3 · edited Mar 02 '17 at 04:29

1

Use mysqli_* instead of mysql_* in your query.

$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$siteSettings = mysqli_query($con,"select * from pu_settings where sid = '1' ");
$siteSoft = mysqli_query($con,"select * from pu_softwares where slug = '$slug' ");

edited Mar 02 '17 at 04:29

Alive to die - Anant

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answered Jul 03 '15 at 07:28

Cyclotron3x3

2,188
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Is it right what you wrote because my knowledge says that mysqli_query need two parameters that is connection object and query. You are giving query only, where is connection object? – Alive to die - Anant Jul 03 '15 at 07:36
temporal blindness..:) – Cyclotron3x3 Jul 03 '15 at 07:40
You could also use the oop style, which is more convenient if you're firing more queries then only one: `$mysqli = new mysqli($host, $user, $pass, $db); $result = $mysqli->query('SELECT * FROM world WHERE people='awesome')` – giorgio Jul 03 '15 at 07:44

Error with this php code

3 Answers3