Replace several texts in PHP include function

Question

I would like to replace several text in a PHP file before I include it in my index.php

For example. The file that I'm about to include in my index.php is a template file called "main.php" and it contains code something like below:

<html>
<head>
<script src="js/jquery.min.js"></script>
<script src="js/jquery-migrate.min.js"></script>
</head>
<body>
Hi, [your_name]. Welcome to [company_name]. Enjoy your stay here!
</body>
</html>

and on my index.php I will have a bunch of queries to get data from database and put it into main.php file. Example below:

<?php
$sql_stmt = "select name, company FROM comp_info WHERE deleted='0' AND status='2'";
$sql = mysqli_query($conn, $sql_stmt) or die(mysqli_error($conn));
$row = mysqli_fetch_assoc($sql);

include_once("main.php");
?>

I tried to use this method below, but I dont know how to use it in this case:

str_replace("[your_name]", $row['name'], main.php);
str_replace("[company_name]", $row['company'], main.php);

So far, what I've searched from the internet is this : Replacing {{string}} within php file But it doesn't works for this case.

Steve · Accepted Answer · 2016-03-08 10:21:07Z

Why reinvent the wheel?

Either pick an existing template engine, or just use vanilla php:

<?php
$sql_stmt = "select name, company FROM comp_info WHERE deleted='0' AND status='2'";
$sql = mysqli_query($conn, $sql_stmt) or die(mysqli_error($conn));
$row = mysqli_fetch_assoc($sql);

$name = $row['name'];
$company = $row['company'];

include_once("main.php");
?>

Template (main.php):

<html>
<head>
<script src="js/jquery.min.js"></script>
<script src="js/jquery-migrate.min.js"></script>
</head>
<body>
Hi, <?=$name?>. Welcome to <?=$company?>. Enjoy your stay here!
</body>
</html>

If you really want to create your own half baked template system, then you would need to use output buffering to capture the contents of the include in a string, then perform your replacements:

<?php
$sql_stmt = "select name, company FROM comp_info WHERE deleted='0' AND status='2'";
$sql = mysqli_query($conn, $sql_stmt) or die(mysqli_error($conn));
$row = mysqli_fetch_assoc($sql);


ob_start();
include_once("main.php");
$mainString = ob_get_clean();

str_replace("[your_name]", $row['name'], $mainString);
str_replace("[company_name]", $row['company'], $mainString);

echo $mainString;
?>

Om R Kattimani · Accepted Answer · 2016-03-08 10:11:53Z

0

create a sting called {%name%} store into content were ever you want then you replace with {%name%} with actual value when you trying to display.

answered Mar 8, 2016 at 10:11

Om R Kattimani

1151 silver badge9 bronze badges

Comments

Karthick Kumar · Accepted Answer · 2016-03-08 10:30:17Z

0

try this

 <?php
$sql_stmt = "select name, company FROM comp_info WHERE deleted='0' AND status='2'";
$sql = mysqli_query($conn, $sql_stmt) or die(mysqli_error($conn));
$row = mysqli_fetch_assoc($sql);


ob_start();
include_once("main.php");
$mainString = ob_get_clean();

strip_tags($row['name']);
strip_tags($row['company']);

echo $mainString;
?>

answered Mar 8, 2016 at 10:30

Karthick Kumar

771 silver badge12 bronze badges

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