PHP str_replace in while loop

Question

The while function in str_replace only returns the first line of information. This is email template.

// I have a macro in my message. (member name)
$message = "Hi {%write_member_name%} how are you?";

$connect = mysql_query("SELECT * FROM members"); 
while($row=mysql_fetch_array($connect)){


$membername = $row["member_name"];

$message = str_replace('{%write_member_name%}', $membername ,$message);

echo $message."<br/>";

}

The result I should have.

Hi Jack how are you?

Hi Mike how are you?

Hi Alex how are you?

But I get the wrong result:

Hi Jack how are you?

If I do not use the str_replace function it works correctly.

How do I use the str_replace function in while loop? Thank you.

It will take a lot of time to fix all the software. But thank you for your suggestion. — Mert Aşan
– Mert Aşan, Commented Apr 14, 2017 at 5:53
@MertA. Have you tried to print the $row["member_name"]? Is it also printing the same value all the time? — Parantap Parashar
– Parantap Parashar, Commented Apr 14, 2017 at 6:00
@Difster If I do not use the str_replace function it works correctly. — Mert Aşan
– Mert Aşan, Commented Apr 14, 2017 at 6:05

Sugumar Venkatesan · Accepted Answer · 2017-04-14 06:06:55Z

2

    $connect = mysql_query("SELECT * FROM members"); 
    while($row=mysql_fetch_assoc($connect)){
        $message = "Hi {%write_member_name%} how are you?";
        $membername = $row["member_name"];

        $message = str_replace('{%write_member_name%}', $membername ,$message);

        echo $message."<br/>";

    }

edited Apr 14, 2017 at 6:06

answered Apr 14, 2017 at 5:52

Sugumar Venkatesan

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4 Comments

Mert Aşan Over a year ago

Sorry this is not the problem. There was a mistake when you were rewriting the codes here.

Sugumar Venkatesan Over a year ago

@Mert A. check it now

Sugumar Venkatesan Over a year ago

use @Stefan Avramovic solution

Sugumar Venkatesan Over a year ago

@Mert A. check my answer

Saty · Accepted Answer · 2017-04-14 06:06:32Z

1

You use sprintf() for this

Replace the percent (%) sign by a variable passed as an argument:

$membername = "";
$connect = mysql_query("SELECT * FROM members");
while ($row = mysql_fetch_array($connect)) {
    $membername = $row["member_name"];
    $message = sprintf("Hi %s how are you?", $membername); // pass argument here
    echo $message . "<br/>";
    $membername = "";
}

mysql_* functions officially deprecated and removed in PHP 7.0.0. You should update your code with PDO or MySQLi.

answered Apr 14, 2017 at 6:06

Saty

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Comments

Stefan Avramovic · Accepted Answer · 2017-04-14 05:58:04Z

0

Much easier:

$connect = mysql_query("SELECT * FROM members"); 
while($icerik=mysql_fetch_array($connect)){

echo $message =  "Hi" . $row["member_name"] . "how are you? <br/>";


}

answered Apr 14, 2017 at 5:58

Stefan Avramovic

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