[英]Jquery, php ajax post 500 Internal Server Error returns
[英]JQuery, Ajax, PHP post error 500 (Internal Server Error)
我使用ajax將數據發布到PHP,此代碼在本地服務器上運行良好(我使用XAMPP),但是當我將其上傳到Web服務器時,它沒有獲取數據並向我顯示此錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
如果我將 dataType 從json更改為application/json並添加contentType: 'application/json; 字符集=utf-8',
那么這是另一個錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
但在本地機器上它工作正常,我可以使用相同的代碼登錄
這是我的 JQuery、AJAX 代碼
$.ajax({
dataType: "json",
type: "POST",
data: {user: sEmail, pass: sPass},
url: "login.php",
success:function(response){
var a = response;
if(a.res=='agree'){
var st = a.role+","+a.name;
$.cookie("imgup", st, { path: '/imgup', expires: 2 });
window.location.href = "logged.php";
}else if (a.res == 'error') {
errorDisplay(a.error);
}
}
});
這是 PHP
<?php
header('Content-Type: application/json');
require_once('inc/init.php');
require_once "helper/helper.php";
$email = sanitize($_POST['user']);
$pass = sanitize($_POST['pass']);
$sql_query = $db->query("SELECT * FROM `user_admin` WHERE `username`='$email'");
if(mysqli_num_rows($sql_query) > 0){
$sql = mysqli_fetch_assoc($sql_query);
if(password_verify($pass, $sql['password'])){
$role = explode(',',$sql['role']);
if(in_array("admin", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'admin','name' => $sql['id']));
}elseif(in_array("editor", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'editor','name' => $sql['id']));
}elseif(in_array("user", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'user','name' => $sql['id']));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_pass'));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_not_exist'));
}
?>
錯誤日志是
PHP 致命錯誤:在第 11 行的 /var/www/html/login.php 中調用未定義的函數 password_verify(),引用: http://__my_IP__/login_page.php
但該功能在本地主機上運行良好!!!
我該如何解決這個錯誤?
注意: sanitize 是檢查字符串中存在的 html 實體的函數
我使用ajax將數據發布到PHP,此代碼在本地服務器上運行良好(我使用XAMPP),但是當我將其上傳到Web服務器時,它沒有獲取數據並向我顯示此錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
如果我將 dataType 從json更改為application/json並添加contentType: 'application/json; 字符集=utf-8',
那么這是另一個錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
但在本地機器上它工作正常,我可以使用相同的代碼登錄
這是我的 JQuery、AJAX 代碼
$.ajax({
dataType: "json",
type: "POST",
data: {user: sEmail, pass: sPass},
url: "login.php",
success:function(response){
var a = response;
if(a.res=='agree'){
var st = a.role+","+a.name;
$.cookie("imgup", st, { path: '/imgup', expires: 2 });
window.location.href = "logged.php";
}else if (a.res == 'error') {
errorDisplay(a.error);
}
}
});
這是 PHP
<?php
header('Content-Type: application/json');
require_once('inc/init.php');
require_once "helper/helper.php";
$email = sanitize($_POST['user']);
$pass = sanitize($_POST['pass']);
$sql_query = $db->query("SELECT * FROM `user_admin` WHERE `username`='$email'");
if(mysqli_num_rows($sql_query) > 0){
$sql = mysqli_fetch_assoc($sql_query);
if(password_verify($pass, $sql['password'])){
$role = explode(',',$sql['role']);
if(in_array("admin", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'admin','name' => $sql['id']));
}elseif(in_array("editor", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'editor','name' => $sql['id']));
}elseif(in_array("user", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'user','name' => $sql['id']));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_pass'));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_not_exist'));
}
?>
錯誤日志是
PHP 致命錯誤:在第 11 行的 /var/www/html/login.php 中調用未定義的函數 password_verify(),引用: http://__my_IP__/login_page.php
但該功能在本地主機上運行良好!!!
我該如何解決這個錯誤?
注意: sanitize 是檢查字符串中存在的 html 實體的函數
我使用ajax將數據發布到PHP,此代碼在本地服務器上運行良好(我使用XAMPP),但是當我將其上傳到Web服務器時,它沒有獲取數據並向我顯示此錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
如果我將 dataType 從json更改為application/json並添加contentType: 'application/json; 字符集=utf-8',
那么這是另一個錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
但在本地機器上它工作正常,我可以使用相同的代碼登錄
這是我的 JQuery、AJAX 代碼
$.ajax({
dataType: "json",
type: "POST",
data: {user: sEmail, pass: sPass},
url: "login.php",
success:function(response){
var a = response;
if(a.res=='agree'){
var st = a.role+","+a.name;
$.cookie("imgup", st, { path: '/imgup', expires: 2 });
window.location.href = "logged.php";
}else if (a.res == 'error') {
errorDisplay(a.error);
}
}
});
這是 PHP
<?php
header('Content-Type: application/json');
require_once('inc/init.php');
require_once "helper/helper.php";
$email = sanitize($_POST['user']);
$pass = sanitize($_POST['pass']);
$sql_query = $db->query("SELECT * FROM `user_admin` WHERE `username`='$email'");
if(mysqli_num_rows($sql_query) > 0){
$sql = mysqli_fetch_assoc($sql_query);
if(password_verify($pass, $sql['password'])){
$role = explode(',',$sql['role']);
if(in_array("admin", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'admin','name' => $sql['id']));
}elseif(in_array("editor", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'editor','name' => $sql['id']));
}elseif(in_array("user", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'user','name' => $sql['id']));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_pass'));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_not_exist'));
}
?>
錯誤日志是
PHP 致命錯誤:在第 11 行的 /var/www/html/login.php 中調用未定義的函數 password_verify(),引用: http://__my_IP__/login_page.php
但該功能在本地主機上運行良好!!!
我該如何解決這個錯誤?
注意: sanitize 是檢查字符串中存在的 html 實體的函數
我使用ajax將數據發布到PHP,此代碼在本地服務器上運行良好(我使用XAMPP),但是當我將其上傳到Web服務器時,它沒有獲取數據並向我顯示此錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
如果我將 dataType 從json更改為application/json並添加contentType: 'application/json; 字符集=utf-8',
那么這是另一個錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
但在本地機器上它工作正常,我可以使用相同的代碼登錄
這是我的 JQuery、AJAX 代碼
$.ajax({
dataType: "json",
type: "POST",
data: {user: sEmail, pass: sPass},
url: "login.php",
success:function(response){
var a = response;
if(a.res=='agree'){
var st = a.role+","+a.name;
$.cookie("imgup", st, { path: '/imgup', expires: 2 });
window.location.href = "logged.php";
}else if (a.res == 'error') {
errorDisplay(a.error);
}
}
});
這是 PHP
<?php
header('Content-Type: application/json');
require_once('inc/init.php');
require_once "helper/helper.php";
$email = sanitize($_POST['user']);
$pass = sanitize($_POST['pass']);
$sql_query = $db->query("SELECT * FROM `user_admin` WHERE `username`='$email'");
if(mysqli_num_rows($sql_query) > 0){
$sql = mysqli_fetch_assoc($sql_query);
if(password_verify($pass, $sql['password'])){
$role = explode(',',$sql['role']);
if(in_array("admin", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'admin','name' => $sql['id']));
}elseif(in_array("editor", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'editor','name' => $sql['id']));
}elseif(in_array("user", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'user','name' => $sql['id']));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_pass'));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_not_exist'));
}
?>
錯誤日志是
PHP 致命錯誤:在第 11 行的 /var/www/html/login.php 中調用未定義的函數 password_verify(),引用: http://__my_IP__/login_page.php
但該功能在本地主機上運行良好!!!
我該如何解決這個錯誤?
注意: sanitize 是檢查字符串中存在的 html 實體的函數
我使用ajax將數據發布到PHP,此代碼在本地服務器上運行良好(我使用XAMPP),但是當我將其上傳到Web服務器時,它沒有獲取數據並向我顯示此錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
如果我將 dataType 從json更改為application/json並添加contentType: 'application/json; 字符集=utf-8',
那么這是另一個錯誤
POST http:// my_server_IP /login.php 500(內部服務器錯誤)
但在本地機器上它工作正常,我可以使用相同的代碼登錄
這是我的 JQuery、AJAX 代碼
$.ajax({
dataType: "json",
type: "POST",
data: {user: sEmail, pass: sPass},
url: "login.php",
success:function(response){
var a = response;
if(a.res=='agree'){
var st = a.role+","+a.name;
$.cookie("imgup", st, { path: '/imgup', expires: 2 });
window.location.href = "logged.php";
}else if (a.res == 'error') {
errorDisplay(a.error);
}
}
});
這是 PHP
<?php
header('Content-Type: application/json');
require_once('inc/init.php');
require_once "helper/helper.php";
$email = sanitize($_POST['user']);
$pass = sanitize($_POST['pass']);
$sql_query = $db->query("SELECT * FROM `user_admin` WHERE `username`='$email'");
if(mysqli_num_rows($sql_query) > 0){
$sql = mysqli_fetch_assoc($sql_query);
if(password_verify($pass, $sql['password'])){
$role = explode(',',$sql['role']);
if(in_array("admin", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'admin','name' => $sql['id']));
}elseif(in_array("editor", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'editor','name' => $sql['id']));
}elseif(in_array("user", $role)){
echo json_encode(array('res' => 'agree', 'role' => 'user','name' => $sql['id']));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_pass'));
}
}else{
echo json_encode(array('res' => 'error', 'error' => 'user_not_exist'));
}
?>
錯誤日志是
PHP 致命錯誤:在第 11 行的 /var/www/html/login.php 中調用未定義的函數 password_verify(),引用: http://__my_IP__/login_page.php
但該功能在本地主機上運行良好!!!
我該如何解決這個錯誤?
注意: sanitize 是檢查字符串中存在的 html 實體的函數
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