How to validate if data already exist in php

Question

Here is code whats wrong with it not working Note:: No error showing or redirect to print.php

$conn=mysqli_connect("localhost","root","","degree");
$sql="select * from students where regno=$regno";
$res= mysqli_query($conn,$sql);
if(mysqli_num_rows($res) >0)
 {
    echo "<script>alert('Student Already Exists')</script)" ;


 }
 else{

    $query="INSERT into students (id,name,regno,program,fname,stcnic,phone,mobile,email,address,gender,gown,relname1,
    relation1,relcnic1,relname2,relation2,relcnic2,gust1,guestrel1,guestcnic1,guest2,guestrel2,guestcnic2) 
     values('','$name',$regno,'$program','$fname','$stcnic',$phone,$mobile,'$email','$address','$gender','$gown','$relname1',
    '$relation1','$relcnic1','$relname2','$relation2','$relcnic2','$guest1','$guestrel1','$guestcnic1','$guest2','$guestrel2','$guestcnic2')";

    $result=mysqli_query($conn,$query);
    if($result)
    {
        $_SESSION['id']=$regno;
        header("location:print.php");
    }
 }

There are no quotes around `$phone,$mobile`. Maybe those aren't just numbers? — KIKO Software, Mar 09 '20 at 07:39
Your query is not prepare (SQL inject approved.), second id with value ' ' is not correct, you must set ID in mysql table autoincrement and don't set in the query. — Simone Rossaini, Mar 09 '20 at 07:55
What happens if another session inserts while this session is checking? — Strawberry, Mar 09 '20 at 07:57
its working fine but ( Alert dosn't working with javascript) -- simple echo"User already exist "; is working i want to show alert message. — Naseeb Ul Hassan, Mar 09 '20 at 08:46

Dj aman motihari · Answer 1 · 2020-03-16T07:08:47.353

You Query Is Wrong :-

$sql="select * from students where regno=$regno";

Correct :-

$sql="select * from students where regno='$regno'";

Second Query

$query="INSERT into students (id,name,regno,program,fname,stcnic,phone,mobile,email,address,gender,gown,relname1,relation1,relcnic1,relname2,relation2,relcnic2,gust1,guestrel1,guestcnic1,guest2,guestrel2,guestcnic2) values('','$name','$regno','$program','$fname','$stcnic','$phone','$mobile','$email','$address','$gender','$gown','$relname1','$relation1','$relcnic1','$relname2','$relation2','$relcnic2','$guest1','$guestrel1','$guestcnic1','$guest2','$guestrel2','$guestcnic2')";

same nothing happened – Naseeb Ul Hassan Mar 09 '20 at 08:11 — Naseeb Ul Hassan, Mar 09 '20 at 08:11

How to validate if data already exist in php

1 Answers1