我用PHP做错了什么 [英] What I'm doing wrong with that PHP
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
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 < span class =code-attribute> 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 ['' < span class =code-string> 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IT屋！
查看全文
