Question

我无法以CRUD网络应用程序的表格形式更新现有数据。查询有什么问题吗？这是我的参考资料来源，我完全按照此处INSERT, UPDATE and DELETE with mysqli执行UPDATE查询。这是我的代码。

<?php
//error_reporting(E_ALL^E_NOTICE);

    function chgDate($date){
    $temp=explode("-",$date);

    return $temp[2]."-".$temp[1]."-".$temp[0];
}

$json=array();


$ic = $_POST['IC'];
$Fic = $_POST['fromIC'];
$name = $_POST['formName'];
$tel = $_POST['formTelephone'];
$gender = $_POST['formGender'];
$email = $_POST['formEmail'];
if(isset($_POST['formUni'])){
    $uni = $_POST['formUni'];
    }
$age = $_POST['formAge'];
$address = $_POST['formAddress'];
$dob = $_POST['formDOB'];
$process= $_POST['process'];

//include ("connect_db.php");
//include_once('connect_db.php');
$db = mysqli_connect("localhost","root","admin","li") or die("Connection Error: " . mysqli_error());

if($process == 'save'){
    $SQL="Insert into biodata (IC, Name, Telephone, Gender, Email, University, Age, Address, DOB) values ('$Fic', '$name', '$tel', '$gender', '$email', '$uni', '$age', '$address', '".chgDate ($dob)."')";
    $json['newrow']=$Fic;
} else if ($process == 'edit') {
    $SQL="UPDATE biodata SET IC='$Fic', Name='$name',  Telephone='$tel', Gender='$gender', Email='$email', University='$uni', Age='$age', Address='$address, DOB ='".chgDate ($dob)."' WHERE IC= '$ic'";
} else if ($process == 'delete') {
    $SQL = "DELETE FROM biodata WHERE IC='$ic'";
}

$data = mysqli_query($db, $SQL);

if($data){
    $json['msg']='success';
}else{
    $json['msg']='fail';
}

echo json_encode($json);

＆GT;

Answer 1

似乎你忘了结束引号

Address='$address'

检查

$SQL="UPDATE biodata SET IC='$Fic', Name='$name',  
Telephone='$tel', Gender='$gender', Email='$email', University='$uni', 
Age='$age', Address='$address', DOB ='".chgDate ($dob)."' WHERE IC= '$ic'";

mysqli update Query有什么问题吗？

1 个答案: