Compare first 4 characters of a string PHP

Question

The problem is with this line:

if $var LIKE '1800%';

and I'm not sure how to fix it. Thanks.

<?php 
//check to see if account number is like 1800*
if (isset($_POST['acct_number'])) {
$var = $_POST['acct_number'];
    if $var LIKE '1800%'; {
        //stop the code
        exit;
    } else {
        echo 'normal account number';
    }
}
?>

Does this answer your question? [How to check if a string starts with a specified string?](https://stackoverflow.com/questions/2790899/how-to-check-if-a-string-starts-with-a-specified-string) — KHansen, Sep 18 '20 at 05:23
Does this answer your question? [startsWith() and endsWith() functions in PHP](https://stackoverflow.com/questions/834303/startswith-and-endswith-functions-in-php) — Wibble, Sep 18 '20 at 09:58

score 2 · Answer 1 · answered Sep 17 '20 at 19:29

2

Another way could be to use strpos()

if (strpos($var, '1800') === 0) {
   // var starts with '1800'
}

answered Sep 17 '20 at 19:29

Ambrish Pathak

3,813
2
15
30

Yes ... perfect! Thank you. – Blondie Sep 22 '20 at 18:38
@Blondie mark any of the answers as accepted whichever is working perfectly for you. – Ambrish Pathak Sep 22 '20 at 19:18

score 1 · Answer 2 · answered Sep 17 '20 at 19:25

1

Use substr function to get first 4 characters and compare it with 1800.

if(substr($var, 0, 4) == '1800')
{
 // your code goes here.
}
``

answered Sep 17 '20 at 19:25

Dark Knight

6,116
1
15
37

score 1 · Accepted Answer · answered Sep 17 '20 at 19:29

1

You need PHP not MySQL. For 1800% just check that it is found at position 0:

if(strpos($var, '1800') === 0) {
    //stop the code
    exit;
} else {
    echo 'normal account number';
}

If it can occur anywhere like %1800% then:

if(strpos($var, '1800') !== false) {
    //stop the code
    exit;
} else {
    echo 'normal account number';
}

answered Sep 17 '20 at 19:29

AbraCadaver

78,200
7
66
87

This worked perfectly! Thank you! if(strpos($var, '1800') === 0) { – Blondie Sep 22 '20 at 18:37

score 1 · Answer 4 · answered Sep 17 '20 at 20:02

1

I would use a regular expression for this preg_match('/^1800.+/', $search, $matches);

answered Sep 17 '20 at 20:02

Ismael Moral

722
1
9
35

May want to use ? instead of + just incase its a 4 char string "1800". – Alex Barker Sep 17 '20 at 20:34

Compare first 4 characters of a string PHP

4 Answers4