Question

嗨，大家好，我在使用PHP和Ajax將我的值插入到MySQL中時遇到了麻煩。我試圖用多個輸入單按鈕插入MySQL。

JS：

$('#save_grade_button').click(function(){ 


       var B_grade = []; 
       var G_grade = []; 


       $.each($("input[id^='student_grde_B']"), function(i, item) { 

       var grade_B = $(item).val(); 

       var stud_id_B = $(item).attr('rel'); 

       B_grade.push({"studnt_B_id":stud_id_B,"studnt_grade_B":grade_B}); 

        }); 

        $.each($("input[id^='student_grde_G']"), function(i, item) { 

         var grade_G = $(item).val(); 

         var stud_id_G = $(item).attr('rel'); 

         G_grade.push({"studnt_G_id":stud_id_G,"studnt_grade_G":grade_G}); 
        }); 

        subjct_id = $('#subjects-list').val(); 
         year_grade_level = $('#year_grade_lvl').val(); 
         sbjct_handler_id = $('#assign-handler-id').val(); 

         $.ajax({ 

          type:'POST', 
          url:'grades.php', 
          dataType:'json', 
          data:{'swtch_numbr':'1','subject_id':subjct_id,'year_level':year_grade_level,'subject_handler_id':sbjct_handler_id,'student_grades_boy':B_grade,'student_grades_girl':G_grade}, 
          success:function (data){ 


          } 
         }); 


       }); 

PHP：

<?php 
include_once('DBconnect.php'); 



$switch_num = $_POST['swtch_numbr']; 


switch ($switch_num) { 
    case 1:   
     save_grades(); 
     break; 


} 
function save_grades(){ 
    session_start(); 

    $school_id = $_SESSION['school_id']; 
    $faculty_id = $_SESSION['user_id_fac']; 
    $subject_id = $_POST['subject_id']; 
    $year_grade_level = $_POST['year_level']; 
    $subject_handeler_id = $_POST['subject_handler_id']; 
    $student_grades_boy = $_POST['student_grades_boy']; 
    $student_grades_girl = $_POST['student_grades_girl']; 


    $save_grades_boy = "INSERT INTO registrar_grade_archive 
            (registrar_archive_id, 
            school_id, 
            subject_id, 
            grade, 
            advisory_faculty_id, 
            subject_handler_id, 
            year_level, 
            student_id) VALUES"; 

        $values_boy = array(); 
        $values_girl = array(); 

        foreach ($student_grades_boy as $key=>$data) { 
           $student_id_B= $data['studnt_B_id']; 
           $grade_B = $data['studnt_grade_B']; 

    $values_boy[$key] = '(\''.$school_id.'\', \''.$subject_id.'\',\''.$grade_B.'\',\''.$faculty_id.'\',\''.$subject_handeler_id.'\',\''.$year_grade_level.'\',\''.$student_id_B.'\')'; 
          } 

     $values_boy = implode(', ', $values_boy); 

    $ready_save_grades_boy .= $values_boy; 


        if(@!mysql_query($ready_save_grades_boy)){ 
         die('error insert'.mysql_error()); 

         } 

} 

>

但是有當我在Firebug的檢查錯誤說： 「你在你的SQL語法錯誤;」

我找不到我做錯了什麼..請幫助，夥計們，因爲我只是Ajax和PHP中的新手。

Answer 1

看起來您開始將SQL存儲在變量$ save_grades_boy中，然後嘗試將值連接到$ ready_save_grades_boy，該變量尚不存在，因此被視爲空字符串。

更換

$ready_save_grades_boy .= $values_boy; 

隨着

$save_grades_boy .= $values_boy; 

或者

$ready_save_grades_boy = $save_grades_boy . $values_boy;