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preg_replace 替换值有子表达式值加数值问题

WBOY
Release： 2016-06-23 13:59:50
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如下面简单的替换：
$ih=100;
$aa='a123b';
$aa=preg_replace('/^(a)123(b)$/i','$1'.$ih.'$2',$aa);
print_r($aa);
怎么结果就是不对，$ih变为字母就正常，数字的话会丢第一位和$1。在$1随便加个字母也正常，不会是我电脑问题吧？


回复讨论(解决方案)

$aa=preg_replace('/^(a)123(b)$/ie','"$1".$ih."$2"',$aa);
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preg_replace('/^(a)*(b)$/i','$1'.$ih.'$2',$aa); 

$ih=500;$aa='a123b';$aa=preg_replace('/^(a)123(b)$/i','${1}'.$ih.'${2}',$aa);print_r($aa);
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'$1'.$ih.'$2'
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相当于 
'$1'.‘100’.'$2'$1100$2
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修改规则 '$1'.$ih.'$2'
实际传递给 preg_replace 的是 ‘$1100$2'
于是 $1 和 $11 就产生了歧义
所以需要人工将其区别开来 ‘${1}100$2'
按你的格式就是 '${1}'.$ih.'$2'
注意：不能使用双引号

都回答的很好，可惜分不多。早知道早点来这问了，郁闷了一整天。

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preg_replace 替换值有子表达式值加数值问题
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