Question

我想存储用户的联系号码。从SQL数据库中的Android注册表单，但当我尝试将$ _post ['contact']代码添加到我的PHP文件时，它显示'失败'而不是'成功'。这是我的PHP文件。

register.php

<?php
 define('HOST','mysql8.000webhost.com');
 define('USER','a6293046_******');
 define('PASS','*********');
 define('DB','a6293046_insti');
 $con = mysqli_connect(HOST,USER,PASS,DB);

 $name = $_POST['name'];
 $email = $_POST['email'];
 $address = $_POST['address'];
 $contact =(int)$_POST['contact'];
 $institute = $_POST['institute'];


 $sql = "insert into Persons (name,email,address,contact-no,institute)      values ('$name','$email','$address','contact-no','$institute')";
 if(mysqli_query($con,$sql)){
 echo 'success';
 }
 else{
 echo 'failure';
 }
 mysqli_close($con);
 ?>

请告诉我我做错了什么。谢谢

Answer 1

    <?php
     define('HOST','mysql8.000webhost.com');
     define('USER','a6293046_******');
     define('PASS','*********');
     define('DB','a6293046_insti');
     $con = mysqli_connect(HOST,USER,PASS,DB);

     $name = $_POST['name'];
     $email = $_POST['email'];
     $address = $_POST['address'];
     $contact =$_POST['contact'];
     $institute = $_POST['institute'];


     $sql = "insert into Persons (name,email,address,contact-no,institute)      values ('$name','$email','$address',$contact,'$institute')";
     $result=$con->query($sql);
if($result)

{
     echo 'success';
     }
     else{
     echo 'failure';
     }
     $con->close();
     ?>



remove the int for $POST_['contact'] and dnt use single quotes('') for $contact while inserting since it is a integer for string value you should give single quotes

Answer 2

试试这个

 <?php
 define('HOST','mysql8.000webhost.com');
 define('USER','a6293046_******');
 define('PASS','*********');
 define('DB','a6293046_insti');
 $con = mysqli_connect(HOST,USER,PASS,DB);
 //change 'somename' with the name of submit button!
 if(isset($_POST['somename'])){
     $error=0;
     $name = $_POST['name'];
     $email = $_POST['email'];
     $address = $_POST['address'];
     $contact = $_POST['contact'];
     $institute = $_POST['institute'];
     if(is_numeric($contact)){
         $error=1;
     }
     if($error==1){
         $sql = "insert into Persons (name,email,address,contact-no,institute) values ('$name','$email','$address','$contact','$institute')";
         $insert= mysqli_query($con,$sql);
         echo 'success';
     } 
     else{
         echo 'failure';
     }
 }
 ?>

此代码仅在$ contact为数字时才会在数据库中插入数据，否则将不会插入

Answer 3

'contact-no'不是变量。将'contact-no'更改为$contact

试试这段代码

$sql = "insert into Persons (name,email,address,contact-no,institute)      values ('$name','$email','$address','$contact','$institute')";

Answer 4

Offcoarse会：你的SQL中有错误，你丢失了美元符号，你应该使用$ contact变量而不是$ contact-no

$sql = "insert into Persons (name,email,address,contact-no,institute)      values ('$name','$email','$address','$contact','$institute')";

如何发布联系号码。使用php

4 个答案: