Question

我在javascript中有一个代码，可以将图像加载到弹出窗口中。我需要先检查标签src中是否有图像。如果存在，请添加

popupContent += '<tr><td><img src="'+mypath+'"/></td></tr>';

否则什么都不做

var mypath = "whatever here but loaded on an ajax get call after hitting a button";
var popupContent = '<table>'; 
popupContent += '<tr><td>Other fields</td></tr>';
popupContent += '<tr><td><img src="'+mypath+'"/></td></tr>';
popupContent += '</table>';

Answer 1

尝试类似的东西：

function checkImageExists(imagePath){

   var httpReq = new XMLHttpRequest();
   httpReq.open('HEAD', imagePath, false);
   httpReq.send();

   return httpReq.status != 404;

}

Answer 2

我修改了您的代码并添加了支票。

 var mypath = "whatever here but loaded on an ajax get call after hitting a button";
var popupContent = '<table>'; 
    popupContent += '<tr><td>Other fields</td></tr>';

if (mypath) {
    popupContent += '<tr><td><img src="'+mypath+'"/></td></tr>';
}

    popupContent += '</table>';

Answer 3

尝试这样的事情：如果图像在那里，它将显示图像，否则将显示No image。

var mypath = "whatever here but loaded on an ajax get call after hitting a button";
var hasImg = mypath ? true : false;
var popupContent = '<table>'; 
popupContent += '<tr><td>Other fields</td></tr>';
hasImg ? popupContent += '<tr><td><img src="'+mypath+'"/></td></tr>' : popupContent += '<tr><td><span>No image</span></td></tr>';
popupContent += '</table>';

javascript如果图片不存在则不加载它

3 个答案: