![做错]( "做错")
本文介绍了我用PHP做错了什么的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 这是我的php代码，其中包含我要提交给更新数据库的表单。 <！DOCTYPE html> < html > < head > < title > Chat-v1.0< / title > < / head > < body > < script src = https ：// ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js\"> < / script > < script type = text / javascript > function submitChat（）{ if （form1.uname .value == || form1.msg.value == ）{ alert（ 所有字段都是必需的!!!）; return ; } var uname = form1.uname.value; var msg = form1.msg.value; var xml = new XMLHttpRequest（）; xml.onreadystatechange = function （）{ if （ xml.readyState == 4 && xml.status == 200 ）{ document .getElementById（' chatlogs'） .innerHTML = xml.responseText; } } xml.open（' GET'， ' insert.php？uname =' + uname + ' & msg =' + msg， true ）; xml.send（）; } < / script > < 表单 方法 = POST 名称 = form1 onsubmit = return false; > 输入ChatName：< 输入 type = text name = uname > < br > 消息：< br > < textarea 姓名 = msg > < / textarea > < br > < 输入 名称 = button type = 提交 id = update value = 更新 onclick = submitChat（） action = > < div id = chatlogs > 正在加载聊天记录，请稍候！ ！ < / div > < / form > < / body > < / html > 这是insert.php文件： 不会发生在数据库中。我能做错什么？ 我的尝试： insert.php不起作用。我认为没有从表单中捕获方法。 if（isset（$ _ POST ['' submit']））{ if （$ _SERVER [' REQUEST_METHOD'] == ' POST'）{ echo pleaseeeeeeeee; $ uname = $ _ POST [' uname']; $ msg = $ _ POST [' msg']; $ username = 根; $ password = ; $ conn = new mysqli（ localhost，$ username，$ password）; mysql_select_db（' chatbox'，$ conn）; mysql_query（ INSERT INTO logs（'username'，'msg'）值（$ uname，$ msg ））; $ result = mysql_query（ SELECT * FROM logs ORDER by id DESC）; while （$ exec = mysql_fetch_array（$ result））{＃code ... echo $ exec [' username']。 ：。$ exec [' msg']。 < br>; } } } else { echo 糟糕; } ？> 解决方案 _POST [' submit']））{ if （ _SERVER [' REQUEST_METHOD'] == ' POST'）{ echo pleaseeeeeeeee; uname = this is my php code with the form that I want to submit to update database.<!DOCTYPE html><html><head><title>Chat-v1.0</title></head><body><script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script><script type="text/javascript">function submitChat(){if (form1.uname.value=="" || form1.msg.value=="") {alert("All fields are required!!!");return;}var uname=form1.uname.value;var msg = form1.msg.value;var xml = new XMLHttpRequest();xml.onreadystatechange=function(){if (xml.readyState==4 && xml.status==200) {document.getElementById('chatlogs').innerHTML=xml.responseText;}}xml.open('GET', 'insert.php?uname='+uname+'&msg='+msg,true);xml.send();}</script><form method="POST" name="form1" onsubmit="return false;">Enter ChatName: <input type="text" name="uname"><br>Message Here:<br><textarea name="msg"></textarea><br><input name = "button" type = "submit" id = "update" value = "Update" onclick="submitChat()" action = ""><div id="chatlogs">Loading Chats, Please Wait!!!</div></form></body></html>this is the insert.php file: Its not happen in the Database. What could I do wrong?What I have tried:The insert.php doesn't work. I think that not catch the method from the form.if(isset($_POST['submit'])) {if ($_SERVER['REQUEST_METHOD'] == 'POST') {echo "pleaseeeeeeeee";$uname=$_POST['uname'];$msg=$_POST['msg'];$username = "root";$password = "";$conn = new mysqli("localhost", $username, $password);mysql_select_db('chatbox',$conn);mysql_query("INSERT INTO logs('username', 'msg') Values($uname, $msg)");$result=mysql_query("SELECT * FROM logs ORDER by id DESC");while ($exec=mysql_fetch_array($result)) {# code...echo $exec['username'].":".$exec['msg']."<br>";}}}else{echo "its bad";}?> 解决方案 _POST['submit'])) {if (_SERVER['REQUEST_METHOD'] == 'POST') {echo "pleaseeeeeeeee";uname= 这篇关于我用PHP做错了什么的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！ 08-07 04:06