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How can I get a string that only contains a to z, A to Z, 0 to 9 and some symbols?



You can test your string (let $str) using preg_match:

if(preg_match("/^[a-zA-Z0-9]+$/", $str) == 1) {
    // string only contain the a to z , A to Z, 0 to 9

If you need more symbols you can add them before ]

Saturday, December 3, 2022


preg_replace("[a-zA-Z0-9_-]", "", $var);

wouldn't even replace those characters, except if the input string is exactly the pattern. By using [] as delimiters, they have not the same effect as their would in the expression itself. You could change your delimiter (e.g.: /), or add some more brackets in the pattern:

preg_replace("/[a-zA-Z0-9_-]/", "", $var);    // this works
preg_replace("[[a-zA-Z0-9_-]]", "", $var);    // this too

Now, to negate a pattern in [], you use ^ at the beginning:

preg_replace("/[^a-zA-Z0-9_-]/", "", $var);

You could also have used the insensitive modifier i to match both lowercase (a-z) and uppercase (A-Z):

preg_replace("/[^a-z0-9_-]/i", "", $var);   // same as above
Monday, December 26, 2022

The Arabic regex is:


Actually, ?-? is a subset of this Arabic range, so I think you can remove them from the pattern.

So, in JS it will be


See regex demo

Tuesday, October 11, 2022
if (preg_match ('/[a-zA-Z0-9 ]/', $getname)) {
Sunday, September 25, 2022

For this PHP regex:

$str = preg_replace ( '{(.)1+}', '$1', $str );
$str = preg_replace ( '{[ '-_()]}', '', $str )

In Java:

str = str.replaceAll("(.)\1+", "$1");
str = str.replaceAll("[ '-_\(\)]", "");

I suggest you to provide your input and expected output then you will get better answers on how it can be done in PHP and/or Java.

Sunday, October 9, 2022
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