为什么str_replace返回的数据错误

例：替换英文逗号，空格字符，或者中文逗号
$str = '正确,联系我们 goodsjob，goodsjob';
$reg = array(',',' ','，' );
$strs = 'username like %'.str_replace($reg,'%, or username like %',$str).'%';
返回的数据是：username like %正确%,%, or username like %or%, or username like %username%, or username like %like%, or username like %%联系我们%, or username like %goodsjob%, or username like %goodsjob%，为什么呢？

正确的结果应该是：usernme like %正确%, or username like %联系我们%, or username like %goodsjob%

回复内容：

例：替换英文逗号，空格字符，或者中文逗号
$str = '正确,联系我们 goodsjob，goodsjob';
$reg = array(',',' ','，' );
$strs = 'username like %'.str_replace($reg,'%, or username like %',$str).'%';
返回的数据是：username like %正确%,%, or username like %or%, or username like %username%, or username like %like%, or username like %%联系我们%, or username like %goodsjob%, or username like %goodsjob%，为什么呢？

正确的结果应该是：usernme like %正确%, or username like %联系我们%, or username like %goodsjob%

先把空格替换成其他字符 $str在替换的过程中，=》 %, or username like % ，这里面也存在满足你替换规则的字符串所以会影响到结果

<code>        $str = '正确,联系我们 goodsjob，goodsjob';
        $str = str_replace(' ', ' ', $str);
        $reg = array(',','，'," ");
        $strs = 'username like %'.str_replace($reg,'%, or username like %',$str).'%';
        
        

</code>

注意：由于str_replace（）函数替换左到右，它可能会进行多次替换时替换以前插入的值。