MYSQL Query Error - Check User in Database

Question

I trying to check if a username is already in use in my db but it's not echoing anything (blank page when i run ) , i want to get a true or false response

<?php

 mysql_connect("localhost", "root", "") or die(mysql_error()); 

 mysql_select_db("data1") or die(mysql_error()); 


$usercheck = "john";

    $sanitizeduser= filter_var($usercheck, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW);

    $check = mysql_query("SELECT EXISTS(SELECT 1 FROM users WHERE username = $sanitizeduser)");

    echo $check;

?>

any way to just get a true/false response . It user exists - $check = false ; else $check = true; — John_Nil
– John_Nil, Commented Feb 5, 2014 at 17:38

Krish R · Accepted Answer · 2014-02-05 17:38:08Z

3

You have shuffled with mysqli_query and mysql_ functions

Try this,

$link = mysqli_connect("localhost","root","","data1") or die("Error " . mysqli_error($link));
$usercheck = "john";

$sanitizeduser= filter_var($usercheck, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW);

$result = $link->query("SELECT username FROM users WHERE username = '".$sanitizeduser."' ");

$row_cnt = $result->num_rows;    
if($row_cnt>0){     
    echo "User Exists"; // echo 1;
}else{      
    echo "No User found"; //echo 0;
}

edited Feb 5, 2014 at 17:38

answered Feb 5, 2014 at 17:16

Krish R

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9 Comments

Krish R Over a year ago

jsfiddle -Only to test your JavaScript, CSS, HTML or CoffeeScript, not php!. BTW, you can use phpfiddle

John_Nil Over a year ago

tried the code above - seems to work - any way to check if result is true or false . I tried to echo result - got an error message (cannot convert to string)

Krish R Over a year ago

To retrieve the results, You need use ex: mysqli_fecth_array() function

John_Nil Over a year ago

its not possible to stick with mysql - it has better syntax (edited code in the main post ) - just need a true/false whether the user exists or not

Krish R Over a year ago

Returns FALSE on Query failure and return TRUE for successful queries, it doesn't mean user exist or not!!

|

Community · Accepted Answer · 2023-11-17 20:33:26Z

2

You are mixing mysql and mysqli:

So your problem is just a typo, just change the function name:

<?php

$link = mysqli_connect("localhost", "root", "password","data1"); 

if (!$link){
    // error handling goes here: mysqli_error()
}
$usercheck = "john";

$sanitizeduser= filter_var($usercheck, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW);

$check = mysqli_query($link,"SELECT EXISTS(SELECT 1 FROM users WHERE username = $sanitizeduser) as check");
?>

Also to display the result you can't just use echo $check, you will have to use something like mysqli_fecth_array for example. See the manual

Example:

while($row = mysqli_fecth_array($check)) {
  echo $row["check"] . "<br>";
}

edited Nov 17, 2023 at 20:33

CommunityBot

11 silver badge

answered Feb 5, 2014 at 17:17

Ibu

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5 Comments

John_Nil Over a year ago

Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\folder\checkuser.php on line 5 .

John_Nil Over a year ago

this line ; mysqli_select_db("data1") or die(mysql_error());

Ibu Over a year ago

@John_Nil you are right, see my update, you will have to pass the connection object to your select_db statement. or even better, pass your db name in the connection

John_Nil Over a year ago

Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\xampp\htdocs\folder\checkuser.php on line 12 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\folder\checkuser.php on line 15

Ibu Over a year ago

Ok, i updated my code. the mysqli_query function takes two parameters, the connection object, and the query. Here is what I suggest, When you have an error like this, just look at the function name in the manual.

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