未捕获的referenceError：$ counter未定义(Uncaught referenceError: $counter is not defined)

未捕获的referenceError：$ counter未定义(Uncaught referenceError: $counter is not defined)

我的JS文件中有一个函数，我希望通过一个名为$ counter的PHP变量给出一个值。 我收到一个错误：$ counter未定义。

继承人PHP代码：

<?php $counter = 0; foreach($res as $category){ echo '<div class="category" onmouseover="doChangeFontWeight($counter);" onmouseout="undoChangeFontWeight($counter);">'; echo '<p class="categoryDescription">'.utf8_encode($category['description']).'</p>'; echo '</div>'; $counter++; } ?>

使用Javascript：

function doChangeFontWeight(counter) { "use strict"; document.getElementsByClassName("categoryDescription")[counter].setAttribute("style", "font-weight: 900");}

I have a function in my JS file which i want to give a value by a PHP variable called $counter. I get an error: $counter is not defined.

Heres the PHP-Code:

<?php $counter = 0; foreach($res as $category){ echo '<div class="category" onmouseover="doChangeFontWeight($counter);" onmouseout="undoChangeFontWeight($counter);">'; echo '<p class="categoryDescription">'.utf8_encode($category['description']).'</p>'; echo '</div>'; $counter++; } ?>

Javascript:

function doChangeFontWeight(counter) { "use strict"; document.getElementsByClassName("categoryDescription")[counter].setAttribute("style", "font-weight: 900");}

最满意答案

'<div class="category" onmouseover="doChangeFontWeight('.$counter.');" onmouseout="undoChangeFontWeight('.$counter.');">';

变量不会在PHP中的单引号内传递。

'<div class="category" onmouseover="doChangeFontWeight('.$counter.');" onmouseout="undoChangeFontWeight('.$counter.');">';

Variables aren't passed inside single quotes in PHP.