2017-04-08 83 views

回答

0

請用適當的方式從阿賈克斯將數據傳遞到PHP

function ajaxfunction(parent) 
    { 
     $.ajax({ 
      type: 'GET', 
      url: 'Connection.php', 
      data: {method:'getFaculties', value:parent} 
      success: function(data) { 
       $("#selFaculty").html(data); 
      } 
     }); 
    } 
0

正確的語法是

url: 'Connection.php?faculties='+getFaculties(parent), 

自認爲是查詢參數,給定一個名字了。

0

使用這樣function Declaration was wrong

function ajaxfunction(parent) 
    { 
     $.ajax({ 
      type: 'GET', 
      url: 'Connection.php?getFaculties='+getFaculties(parent), 
      success: function(data) { 
       $("#selFaculty").html(data); 
      } 
     }); 
    } 
0

首先調用你的函數和變量獲得的返回值,然後把你的Ajax請求。

function ajaxfunction(parent) 
{ 
    var data_in = getFaculties(parent); 
    $.ajax({ 
     type: 'GET', 
     url: 'Connection.php?getFaculties='+data_in, 
     success: function(data) { 
      $("#selFaculty").html(data); 
     } 
    }); 
}