Question

有人可以帮助我，我需要将变换字符串作为：这是HTML代码：

<p><img src="/images/blog//023.jpg" alt="sposi a venezia" width="910" height="620" data-src="salvatorefranco.com/images/blog//023.jpg"; /><img src="/images/blog//wedding-photographer-in-portofino.jpg" alt="sposi a venezia" width="910" height="620" data-src="salvatorefranco.com/images/blog//…; /></p>


alt=""   to 
alt="title image"

我试过了：

$var = str_replace('alt=""', 'alt="'.$title_image.'"', $object);

但没有工作...... 谢谢萨尔瓦多

Answer 1

你可以这样做：

var watcher = new FileSystemWatcher();
watcher.Created += (sender, e) =>
{
    var lines = File.ReadAllLines(e.FullPath, Encoding.UTF8); //io exception here
};

Answer 2

您的代码应该可以正常运行。

<?php

$object = '<p><img src="/images/blog/023.jpg" alt="" width="910" height="620" data-src="/images/blog/023.jpg" /><img src="/images/blog//wedding-photographer-in-portofino.jpg" alt="" width="910" height="620" data-src="/blog//wedding-photographer-in-portofino.jpg" /></p>';

$title_image = "title image";

echo $var = str_replace('alt=""', 'alt="'.$title_image.'"', $object);

＆GT;

输出

<p><img src="/images/blog/023.jpg" alt="title image" width="910" height="620" data-src="/images/blog/023.jpg" /><img src="/images/blog//wedding-photographer-in-portofino.jpg" alt="title image" width="910" height="620" data-src="/blog//wedding-photographer-in-portofino.jpg" /></p>

Working Link

Answer 3

我们可以尝试str_replace函数来解决这个问题

 $alt='alt=""' ;
 $var = str_replace("{$alt}", 'alt="title image"',"{$alt}"); 
 echo $var;

Php STR_REPLACE不适用于双引号

3 个答案: