Question

我正在从iphone应用程序向mysql插入数据，但它没有将数据插入表中我回显结果和输入值但它也没有显示任何值

HTML CODE对于页面

  <html>


  <head>
  <title>data to server</title>
  </head>
  <body>
  <form action="surveyAnswer.php" method="post" enctype="multipart/form-data"><br>

  <INPUT TYPE = "Text" VALUE ="1" NAME = "survey_question_response_id">

  <INPUT TYPE = "Text" VALUE ="1" NAME = "survey_id">
  <INPUT TYPE = "Text" VALUE ="1" NAME = "question_id">
  <INPUT TYPE = "Text" VALUE ="1" NAME = "survey_response_answer_id">
   <input type="submit" value="Upload File">
  </form>
  </body>
  </html>



     <?php
     $host = ""; 
     $user = ""; 
     $pass = ""; 
     $database = ""; 

     $linkID = mysql_connect($host, $user, $pass) or die("Could not connect to host."); 
     mysql_select_db($database, $linkID) or die("Could not find database."); 

     $survey_question_response_id=$_POST['survey_question_response_id'];
     $survey_id=$_POST['survey_id'];
     $question_id=$_POST['question_id'];
     $survey_response_answer_id=$_POST['survey_response_answer_id'];
     echo($survey_question_response_id);
     $query=("INSERT INTO survey_question_responses   (survey_question_response_id,survey_id,question_id,survey_response_answer_id)
     VALUES ('$survey_question_response_id', '$survey_id','$question_id','$survey_response_answer_id')");
     mysql_query($query,$con);
     printf("Records inserted: %d\n", mysql_affected_rows());
     echo($survey_id) 
    ?>

Answer 1

您的form方法为POST，您正在使用$_GET来捕获变量。使用$_POST代替$GET来解决您的问题。

另外，标签语法错误。

<form action="surveyAnswer.php" method="post" enctype="multipart/form-data">

这指向surveyAnswer.php。因此，将以下代码放在surveyAnswer.php页面中，并将其从显示html表单的页面中删除。

    <?php

       $survey_question_response_id=$_POST['survey_question_response_id'];
       $survey_id=$_POST['survey_id'];
       $question_id=$_POST['question_id'];
       $survey_response_answer_id=$_POST['survey_response_answer_id'];
       $query=("INSERT INTO survey_question_responses     (survey_question_response_id,survey_id,question_id,survey_response_answer_id)
      VALUES ('$survey_question_response_id', '$survey_id','$question_id','$survey_response_answer_id')");
     mysql_query($query,$con);
     printf("Records inserted: %d\n", mysql_affected_rows());
     echo($survey_id);

 ?>

Answer 2

你用 POST 来拍摄你的动作，但在你的PHP代码中你使用$ _GET

更改所有$ _GET：

 <?php
   $survey_question_response_id=$_POST['survey_question_response_id'];
   // OR eighter to
   $survey_question_response_id=$_REQUEST['survey_question_response_id'];
   ....

Answer 3

在表单中，您有form method =“post”，当您使用GET存储值时，更改表单方法=“GET”或使用

$_POST['survey_question_response_id'];

而不是

$_GET['survey_question_response_id'];

等其他变量也是如此。

你也错过了“;”这里“echo（$ survey_id）”

将数据插入mysql数据库无法正常工作

3 个答案: