php - php:使用php脚本替换图像src路径

我想创建PHP脚本来更新img src路径。

旧结果：

<a href="http://example.com" target="_blank"><img src="{{media url=''}}wysiwyg/images/img_07.jpg" border="1">

预期结果 ：

<a href="http://example.com" target="_blank"><img src="{{media url='wysiwyg/images/img_07.jpg'}}" border="1">

我用下面的代码：

<?php
$servername = "localhost";
$username = "root";
$password = "testing";
$dbname = "database";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT * FROM table_nm WHERE attribute_id = 1";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while ($row = $result->fetch_assoc()) {
        echo htmlentities($row["value"]) . "<br>"; //old result text display from here.
    }
} else {
    echo "0 results";
}
$conn->close();
?>

为了获得预期的结果，我需要改变什么？

最佳答案

您可以尝试这样做以获得预期的结果

$patterns = ["'}}","jpg","png","jpeg","gif"];
$replacements = ["","jpg'}}","png'}}","jpeg'}}","gif'}}"];
$row["value"] = str_replace($patterns, $replacements, $row["value"]);

关于php - php:使用php脚本替换图像src路径，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55902390/