如果添加一个额外的参数,为什么不调用javascript函数

问题描述 投票:0回答:2

我有一个很奇怪的情况,我正在这样调用我的javascript函数...

window.top.window.stopUpload(<? echo $result; ?>,<? echo $file_name; ?>);

JavaScript函数看起来像这样,

function stopUpload(success,filePath){
          var result = '';
          if (success == 1){
             result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>';
          }
          else {
             result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>';
          }
          document.getElementById('f1_upload_process').style.visibility = 'hidden';
          document.getElementById('f1_upload_form').innerHTML = result + '<input name="image_file" type="file" class="browse" /><input type="submit" name="submit_button" value="Upload"  class="browse"/>';
          document.getElementById('f1_upload_form').style.visibility = 'visible';     

          return true;   
    }

以上代码不执行stopUpload功能。


但是,如果我这样做,

window.top.window.stopUpload(<? echo $result; ?>);

和类似的javascript,

function stopUpload(success){
          var result = '';
          if (success == 1){
             result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>';
          }
          else {
             result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>';
          }
          document.getElementById('f1_upload_process').style.visibility = 'hidden';
          document.getElementById('f1_upload_form').innerHTML = result + '<input name="image_file" type="file" class="browse" /><input type="submit" name="submit_button" value="Upload"  class="browse"/>';
          document.getElementById('f1_upload_form').style.visibility = 'visible';     

          return true;   
    }

有了一个参数,它就起作用了!

问题

为什么它只适用于一个参数而不适用于两个参数?我尝试发送像'hello'而不是$file_name这样的普通字符串,但仍然无法调用。

php javascript function parameters parameter-passing
2个回答
3
投票

这样调用您的函数:

window.top.window.stopUpload(<? echo $result; ?>,'<? echo $file_name; ?>');

希望有帮助。


0
投票

尝试一下:

window.top.window.stopUpload('<? echo $result; ?>','<? echo $file_name; ?>');

请记住,$ result不应为任何数值。用户$ result ='1'。

并且在if语句中更改成功=='1')。

希望有帮助。

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