To replace the
with the in php in an array

Question

<?php
  $servername="localhost";
  $username="root";
  $password="";
  $dbname="abc";
  $sql=mysqli_connect($servername,$username,$password,$dbname)
  $get=mysqli_query($sql,"SELECT * FORM table_name")

  $dta=array();
  while($fetch=mysqli_fetch_assoc($get)){
      array_push($dta,$fetch);
  }

  $dta=preg_replace("/<br>/",'\n',$dta);
  $json=json_encode($dta);
  echo $json;
  exit;
?>

I want to replace the data form the data base. Want to replace column values from
with

For example:

I have some columns like A, B, C
Column A has data like qwe qu
I want the result of column A would be like qwe <n> qu

Note: The above small example has more than 20 columns.

There is also a function called `nl2br()` http://php.net/manual/en/function.nl2br.php — JustOnUnderMillions, Apr 20 '17 at 10:44

Hossam · Answer 1 · 2017-04-20T10:55:45.387

0

Change:

$dta=preg_replace("/<br>/",'\n',$dta);

To:

array_walk($dta, function(&$val){
    $val = str_replace("<br>", "\r\n", $val);
});

& keep in mind:

For \n use double quotes " instead of single quotes '.
If the resulted string having new lines will be used in Windows, replace with \r\n instead of just \n
There is no need for preg_replace, str_replace would perfectly fit
As $dta is an array, we have to use array_walk or loop in the array.

edited Apr 20 '17 at 10:55

answered Apr 20 '17 at 10:38

Hossam

1,126
8
19

$dta = an array. you will have to loop and replace. – Niek van der Maaden Apr 20 '17 at 10:41
Your callback does not have an input parameter `function(){`, so within the function you have no varaible with the current value. `$dta` does not exists in the function. – JustOnUnderMillions Apr 20 '17 at 10:46
geting dta is undefined @HossamMagdy – banu prakash Apr 20 '17 at 10:50
Notice: Undefined variable: dta in the line 26 – banu prakash Apr 20 '17 at 10:55
Did you replace the exactly mentioned line as written above ? – Hossam Apr 20 '17 at 10:56
yes,once u replace my totall code with your anwser please – banu prakash Apr 20 '17 at 10:59
yes i replace this line $dta=preg_replace("/
/",'\n',$dta); with your code – banu prakash Apr 20 '17 at 11:00

score 0 · Accepted Answer · edited May 23 '17 at 12:34

0

Your code has so many bugs.

$fetch holds an array (not a string)
$dta is then an array of arrays.
preg_replace works only on strings not arrays.

You have first to fix that issue.

Replace this line

array_push($dta,$fetch);

with this:

//update all columns from the current fetched row
$fetch = array_map(function($a){ 
       return str_replace(['<br/>','<br>','<br />'],"\n",$a);
},$fetch);
//add the data
array_push($dta,$fetch);

and remove the preg_match line.

What does [' ',' ',' '] in str_replace?

If you have an array as first parameter then each entry will searched an replaced. In this example we dont really know how   is written in the code. Can be  , so i have added all the cases here.

More infos:

http://php.net/manual/en/function.str-replace.php use of str_replace

http://php.net/manual/en/function.array-map.php use of array_map

http://php.net/manual/en/functions.anonymous.php use of function(){}

https://stackoverflow.com/a/10304027/4916265 more about function()use(){}

edited May 23 '17 at 12:34

Community

1
1

answered Apr 20 '17 at 10:58

JustOnUnderMillions

3,741
9
12

ok replaced the code but the
is not replaced with the "\n" – banu prakash Apr 20 '17 at 11:05
@banuprakash updated. Please dont only copy&paste, enhance what we give you :-) – JustOnUnderMillions Apr 20 '17 at 11:06
i replaced it in the while loop – banu prakash Apr 20 '17 at 11:08
where to write fetch query – banu prakash Apr 20 '17 at 11:09
i am new to php – banu prakash Apr 20 '17 at 11:09
@banuprakash I updated my question. And if you use mysqli & json you can not be so new to php ;-) One of the best places to get info about functions http://php.net like http://php.net/manual/en/function.str-replace.php – JustOnUnderMillions Apr 20 '17 at 11:12

To replace the with the in php in an array

2 Answers2

To replace the
with the in php in an array