如果语句解析不正确

问题描述

我想把这个if语句作为：
如果第一个字符串位置是.png，然后从大海捞针获得$ png1，但如果第一个字符串位置是.jpg，那么得到$ jpg1从干草堆，但如果它是.gif，从haystack获取$ gif1，否则如果没有找到它们那么字符串位置是.bmp所以得到$ bmp1

I am trying to get this if statement to follow as:if the first string position is .png, then get $png1 from a haystack, but if the first string position is .jpg, then get $jpg1 from the haystack, but if it is .gif, get $gif1 from haystack, else if none of them are found then the string position is .bmp so get $bmp1

这是我试过的，但它没有正确解析：

Here is what i tried, but it doesn't parse correctly:

<?php
// if first occurence is .png get $png1 needle from haystack
if (preg_match('#cid:([^"@]*).png@([^"]*)#', $html_part))
{           $find = '#cid:([^"@]*).png@([^"]*)#';
    $replace1 = $png1;
    $html_part = preg_replace($find, $replace, $html_part);
}
// if first occurence is .jpg get $jpg1 needle from haystack
elseif (preg_match('#cid:([^"@]*).jpg@([^"]*)#', $html_part))
{           $find = '#cid:([^"@]*).jpg@([^"]*)#';
    $replace1 = $jpg1;
    $html_part = preg_replace($find, $replace, $html_part);
}
// if first occurence is .gif then get $gif1 needle from haystack
elseif (preg_match('#cid:([^"@]*).gif@([^"]*)#', $html_part))
{           $find = '#cid:([^"@]*).gif@([^"]*)#';
    $replace = $gif1;
    $html_part = preg_replace($find, $replace, $html_part);
}
// if first occurence is .bmp then get $bmp1 needle from haystack
else
{           $find = '#cid:([^"@]*).bmp@([^"]*)#';
    $replace = $bmp1;
    $html_part = preg_replace($find, $replace, $html_part);
}
?>

示例 $ html_part ，包含换行符添加以供显示，是：

An example $html_part, with line breaks added for display, is:

<b>Bold Text.</b> <i>Italicized Text.</i> <u>Underlined Text.</u> Plain Unformatted Text.
<img width=183 height=183 id="Picture_x0020_3" src="cid:[email protected]"
alt="Description: Description: Description: cid:[email protected]">
<img width=153 height=145 id="Picture_x0020_2" src="cid:[email protected]"
alt="Description: Description: cid:[email protected]"><img width=182 height=123
id="Picture_x0020_1" src="cid:[email protected]"
alt="Description: Description: cid:[email protected]">`

推荐答案

错误在你的代码中只是

    $replace1 = $png1;
    $html_part = preg_replace($find, $replace, $html_part);

和

    $replace1 = $jpg1;
    $html_part = preg_replace($find, $replace, $html_part);

- 你设置 $ replace1 ，但是你使用 $ replace 。

- you set $replace1, but you use $replace.

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