Question

美好的一天！无论出于何种原因，如果数字大于10，请指定str_replace()更改$replace参数，裁减单位并只留下几十个？

输入数据（$data）：

...
<div onclick="window.location.href='/template-04.php?type=users&char=7';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=8';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=9';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=10';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=11';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=12';"></div>
...

非常简单的PHP代码：

for($axx = 0; $axx < 68; $axx ++)
{
    $z = '['.$axx.']';

    $newName = 'templ4-user-'.$z.'.html?'.$z;

    echo '<br>'.$newName; // echo (axx = 13):  <br>templ4-user-[13].html?[13]

    $data = str_replace('template-04.php?type=users&char='.$axx, $newName, $data);
}

结果$data不正确。（if $axx > 10）为什么？

...
<div onclick="window.location.href='/templ4-user-[7].html?[7]';"></div>
<div onclick="window.location.href='/templ4-user-[8].html?[8]';"></div>
<div onclick="window.location.href='/templ4-user-[9].html?[9]';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]0';"></div>   <------ !!!!!!!
<div onclick="window.location.href='/templ4-user-[1].html?[1]1';"></div>   <------ !!!!!!! 
<div onclick="window.location.href='/templ4-user-[1].html?[1]2';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]3';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]4';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]5';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]6';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]7';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]8';"></div>
<div onclick="window.location.href='/templ4-user-[1].html?[1]9';"></div>
<div onclick="window.location.href='/templ4-user-[2].html?[2]0';"></div>
...

请帮忙。

Answer 1

这是因为在第一次迭代中，所有1将变为[1]，这意味着12将变为[1]2并且将永远不再匹配12。

您可以使用preg_replace：

代替循环

$data = <<<EOS
<div onclick="window.location.href='/template-04.php?type=users&char=7';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=8';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=9';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=10';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=11';"></div>
<div onclick="window.location.href='/template-04.php?type=users&char=12';"></div>
EOS;

$pattern = '/template-04.php\?type=users&char=(\d+)/i';
$replacement = 'templ4-user-[$1].html?[$1]';
echo preg_replace($pattern, $replacement, $data);

结果：

<div onclick="window.location.href='/templ4-user-[7].html?[7]';"></div>
<div onclick="window.location.href='/templ4-user-[8].html?[8]';"></div>
<div onclick="window.location.href='/templ4-user-[9].html?[9]';"></div>
<div onclick="window.location.href='/templ4-user-[10].html?[10]';"></div>
<div onclick="window.location.href='/templ4-user-[11].html?[11]';"></div>
<div onclick="window.location.href='/templ4-user-[12].html?[12]';"></div>

str_replace工作不正确（“str_replace”更改$ replace参数）

1 个答案: