Ich habe 1 Tabelle und möchte 2 Werte in MySql einfügen. Es gibt jedoch nur einen Wert in meiner Datenbank und der andere bleibt leer.Daten in MySql mit 2 Eingabewerten einfügen
Das Design meiner mySql Tabelle:
Meine Html-Tabelle Code ist unten:
<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>
Mein post.php Code ist unten:
<?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);
?>
Mein addCreditAssessment.php-Code ist unter:
<?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 "{$query2}";
//header('Location: post.php');
}
}
?>
Sie erhalten '$ name, $ Nationalität, $ age' und legen Sie' '$ nameJoint1', '$ nationJoint1', '$ ageJoint1'' in zweiter Funktion 'addApplicantPersonalDetails2 () ' –
Checkout die Antwort von mir gegeben: [Meine Antwort] (https://stackoverflow.com/questions/44879391/insert-data-to-mysql-with-2-input-value/44879472#44879472) –