Question

我有1个表，我想在MySql中插入2个值。但是，我的数据库中只有一个值，另一个值为空。

mySql表的设计：

我的Html表格代码如下：

<table>

   <tr>
       <th></th>
       <th>Main Applicant</th>
       <th>Joint Applicant1</th>

  </tr>
  <tr>
       <td>Name</td>
       <td><input type="text"   id="nameMain" name="nameMain" class="form-control" autocomplete="off" required></td>
      <td><input type="text"   id="nameJoint1" name="nameJoint1" class="form-control" autocomplete="off" required></td>


  </tr>

  <tr>
      <td>Nationality(Country)</td>
      <td><input type="text"  id="nationMain" name="nationMain" class="form-control" autocomplete="off" required></td>
      <td><input type="text"  id="nationJoint1" name="nationJoint1" class="form-control" autocomplete="off" required></td>

 </tr>
 <tr>
    <td>Age</td>
    <td><input type="number"  id="ageMain" name="ageMain" class="form-control" autocomplete="off" required></td>
    <td><input type="number"  id="ageJoint1" name="ageJoint1" class="form-control" autocomplete="off" required></td>


</tr>

我的post.php代码如下：

       <?php 


    require_once 'db/dbfunction.php';
    require_once 'db/dbCreditAssessment.php';

    session_start();
    $con = open_connection();



    $name = $_POST['nameMain'];
    $nationality = $_POST['nationMain'];
    $age = $_POST['ageMain'];


    $nameJoint1 = $_POST['nameJoint1'];
    $nationJoint1 = $_POST['nationJoint1'];
    $ageJoint1 = $_POST['ageJoint1'];

    addApplicantPersonalDetails($con,$name,$nationality,$age);

    addApplicantPersonalDetails2($con,$name,$nationality,$age);

    close_connection($con);


    ?>

我的addCreditAssessment.php代码如下：

    <?php

function addApplicantPersonalDetails($con,$name,$nationality,$age){

    $query = "insert into zzz(name,nationality,age) 
            values('$name','$nationality','$age')";            
                 //echo "{$sqlString}";



                 $insertResult = mysqli_query($con, $query);


                 if($insertResult){
                     echo " Applicant Detail Added !<br />";
                     echo "<a href='index.php'>Back to Home</a>";
                 }
                 else {
                     echo " Error !";
                     echo "{$query}";
                     //header('Location: post.php');
                 }


}

function addApplicantPersonalDetails2($con,$name,$nationality,$age){



    $query2 = "insert into zzz(name,nationality,age) 
            values('$nameJoint1','$nationJoint1','$ageJoint1')"; 


                 $insertResult2 = mysqli_query($con, $query2);



                 if($insertResult2){
                     echo " Applicant Detail Added !<br />";
                     echo "<a href='index.php'>Back to Home</a>";
                 }
                 else {
                     echo " Error !";
                     echo "{$query2}";
                     //header('Location: post.php');
                 }
}


?>

Answer 1

您正在使用未定义的变量i为空的原因

这样做

function addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1){
    $query2 = "insert into zzz(name,nationality,age) 
            values('$nameJoint1','$nationJoint1','$ageJoint1')"; 


                 $insertResult2 = mysqli_query($con, $query2);



                 if($insertResult2){
                     echo " Applicant Detail Added !<br />";
                     echo "<a href='index.php'>Back to Home</a>";
                 }
                 else {
                     echo " Error !";
                     echo "{$query2}";
                     //header('Location: post.php');
                 }
}

它适用于你。

Answer 2

你的第二个功能是错误的：

您的职能是：

$query2 = "insert into zzz(name,nationality,age) 
            values('$name','$nationality','$age')";

像这样写你的SQL：

 line 21, in <module> average = 
 sum(student_info[student_name]/len(student_info))TypeError: 'float' object 
 is not iterable"

您可以从$ _POST请求设置旧变量，而不是从函数

设置

Answer 3

我认为这里有两个错误。第一个是在这个调用中，你使用与上面相同的参数：

addApplicantPersonalDetails2($con,$name,$nationality,$age);

我猜你的意思是：

addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1);

其次，正如@FrayneKonok和@Richard指出的那样，你在函数体内的查询中使用了错误的形式参数。

Answer 4

您正在接收$name,$nationality,$age并在第二个函数'$nameJoint1','$nationJoint1','$ageJoint1'插入addApplicantPersonalDetails2()。

您需要对该功能使用相同的查询。您的第二个查询应该是：

$query2 = "insert into zzz(name,nationality,age) values('$name','$nationality','$age')";

或者您可以使用'$nameJoint1','$nationJoint1','$ageJoint1'

更改功能参数

Answer 5

您可以使用nameMain作为字段

，而不是使用nameJoint1和name[]

在post.php上，名称用户输入将属于$_POST['name']数组。

这样您就可以使用for循环来提交所有这些数据。

所以你的html文件看起来像这样：

<table>

   <tr>
       <th></th>
       <th>Main Applicant</th>
       <th>Joint Applicant1</th>

  </tr>
  <tr>
       <td>Name</td>
       <td><input type="text"   id="nameMain" name="name[]" class="form-control" autocomplete="off" required></td>
      <td><input type="text"   id="nameJoint1" name="name[]" class="form-control" autocomplete="off" required></td>


  </tr>

  <tr>
      <td>Nationality(Country)</td>
      <td><input type="text"  id="nationMain" name="nation[]" class="form-control" autocomplete="off" required></td>
      <td><input type="text"  id="nationJoint1" name="nation[]" class="form-control" autocomplete="off" required></td>

 </tr>
 <tr>
    <td>Age</td>
    <td><input type="number"  id="ageMain" name="age[]" class="form-control" autocomplete="off" required></td>
    <td><input type="number"  id="ageJoint1" name="age[]" class="form-control" autocomplete="off" required></td>


</tr>

然后您的功能将如下所示：

function addApplicantPersonalDetails($con,$name,$nationality,$age){
if(count($name) == count($nationality) == count($age) = $count){
    for($i = 0; $i < $count; $i++){
       $query = "insert into zzz(name,nationality,age) 
            values('$name[$i]','$nationality[$i]','$age[$i]')";            
                 //echo "{$sqlString}";

                 $insertResult = mysqli_query($con, $query);
    }

                 if($insertResult){
                     echo " Applicant Detail Added !<br />";
                     echo "<a href='index.php'>Back to Home</a>";
                 }
                 else {
                     echo " Error !";
                     echo "{$query}";
                     //header('Location: post.php');
                 }


}

Answer 6

<强> post.php中

更改变量名称

addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1);

<强> addCreditAssessment.php

function addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1)

使用2个输入值将数据插入MySql

6 个答案: