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 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 "{$query2}";
//header('Location: post.php');
}
}
?>
最佳答案
您使用的变量未在函数中定义,为什么为空
这样做
function addApplicantPersonalDetails2($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。
关于php - 使用 2 个输入值将数据插入到 MySql,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44879391/