如何在 ajax 加载后加载脚本?

How to load script after ajax load?

var currentTallest = 0,
 currentRowStart = 0,
 rowDivs = new Array(),
 $el,
 topPosition = 0;

$('.blocks').each(function() {

$el = $(this);
topPosition = $el.position().top;

if (currentRowStart != topPosition) {

 // we just came to a new row.  Set all the heights on the completed row
 for (currentDiv = 0 ; currentDiv < rowDivs.length ; currentDiv++) {
   rowDivs[currentDiv].height(currentTallest);
 }

 // set the variables for the new row
 rowDivs.length = 0; // empty the array
 currentRowStart = topPosition;
 currentTallest = $el.height();
 rowDivs.push($el);

} else {

 // another div on the current row.  Add it to the list and check if it's taller
 rowDivs.push($el);
 currentTallest = (currentTallest < $el.height()) ? ($el.height()) : (currentTallest);

}

// do the last row
for (currentDiv = 0 ; currentDiv < rowDivs.length ; currentDiv++) {
  rowDivs[currentDiv].height(currentTallest);
}

});​

您好,我使用上面的代码为 bootstrap 创建了等高的列,并且有效。但我还使用无限滚动库来加载更多内容,在我向下滚动后脚本不会加载到新列中,我知道这是因为 ajax 在脚本加载后加载。我知道 .ajaxComplete() 句柄应该可以完成这项工作,但我还没有找到正确的方法。

我不确定我是否答对了问题,但您可能需要检查 $.getScript() 函数并在 ajax 调用完成后使用它:

$.ajax({

    // rest of the code

    complete: function() {
         $.getScript("path/to/script.js", function() {
              alert('All done!');
         });
    }
})

希望这对您有所帮助!