Php function do not work with javascript

Question

I have some php and javascript, everythings work and I do not get any error message. I can post data without any error to database. And everythings looks good but I cannot use some function in php.

Example.

I have a textarea and I send its data with ajax to a php file. In php file a need to use str_replace function. I can insert data to database in same php file without any error but the function that I try to use like str_replace or mysqli_real_escape_string, etc. do not work.

What would be the reason?

Here codes.

    $(".editBoxButton").click(function(){

        var yazi = jQuery('#editInput').val();

            $.ajax({
                type: 'POST' ,
                url : 'ajax/editEntry.php',
                    data: {
                    text: yazi,
                    },
                success : function(d){ 
                alert(d);
                location.reload();  //refresh
                }
            });

    });

ajax/editEntry.php

<?php

$yeniyazi=$_POST['text'];   
$yeniyazi = str_replace("\n", "<br>", $yeniyazi);

$s=$yeniyazi;

echo json_encode($s);

?>

in the alert, I get still \n. it does not replaced.

I do not get any error. only str_replace do not work that is my problem. expect str_replace, it works properly.

make sure you use double quotes for special characters as \n (happened to me). I know your answer has double quotes, just in case you didn't copy-paste — bmurauer
– bmurauer, Commented Aug 7, 2013 at 7:17
json_encode turns an array/object into a string. You seem to be working on a string already, so there is no need for a json_encode. — Sumurai8
– Sumurai8, Commented Aug 7, 2013 at 7:22

Nadh · Accepted Answer · 2013-08-07 07:14:43Z

1

You should probably just do:

$yeniyazi = nl2br($yeniyazi);

If you really want to manually replace, use regex.

$yeniyazi = preg_replace("/\r?\n/s", "<br />", $yeniyazi);

answered Aug 7, 2013 at 7:14

Nadh

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1 Comment

ernie Over a year ago

nl2br also has the benefit that it will handle different forms of new lines (i.e. \r\n, \n\r, \n, and \r)

Fabien Papet · Accepted Answer · 2013-08-07 07:15:51Z

1

Actually, you have to escape the \ with another one or use singlequotes instead.

$val = str_replace("\\n" , "<br>", $val);

Or you can do it in a single line like this :

<?php 
echo json_encode(nl2br($_POST['text']));

answered Aug 7, 2013 at 7:15

Fabien Papet

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Comments

Samira Khorshidi · Accepted Answer · 2013-08-07 07:37:58Z

i hope this example help you:

<?php
// Order of replacement
$str     = "Line 1\nLine 2\rLine 3\r\nLine 4\n";
$order   = array("\r\n", "\n", "\r");
$replace = '<br />';

// Processes \r\n's first so they aren't converted twice.
$newstr = str_replace($order, $replace, $str);

 // Outputs F because A is replaced with B, then B is replaced with C, and so on...
// Finally E is replaced with F, because of left to right replacements.
$search  = array('A', 'B', 'C', 'D', 'E');
$replace = array('B', 'C', 'D', 'E', 'F');
$subject = 'A';
echo str_replace($search, $replace, $subject);

// Outputs: apearpearle pear
// For the same reason mentioned above
$letters = array('a', 'p');
$fruit   = array('apple', 'pear');
$text    = 'a p';
$output  = str_replace($letters, $fruit, $text);
echo $output;
?>

Pao Im · Accepted Answer · 2015-12-11 20:09:57Z

0

I also have problem with \n using in javascript and PHP.

Current Env.:

PHP 5.3.10
Javascript
XHTML 1.0

The problem:

....
if (rowcount == 5) {
    openShipments += '\n';
    rowcount = 0;
}
....

It works when using \\n bellow:

....
if (rowcount == 5) {
    openShipments += '\\n';
    rowcount = 0;
}
....

answered Dec 11, 2015 at 20:09

Pao Im

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