php - 使用 2 个输入值将数据插入到 MySql

Question

我有 1 个表，我想将 2 个值插入到 MySql 中。但是，只有一个值进入我的数据库，另一个仍为空。

我的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

您使用的变量未在函数中定义，为什么为空

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');
                 }
}

这对你有用。