info.php<script type="text/javascript" language="javascript">
function InitAjax()
{
var ajax=false;
try {
ajax = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
ajax = new ActiveXObject("Microsoft.XMLHTTP");
} catch (E) {
ajax = false;
}
}
if (!ajax && typeof XMLHttpRequest!='undefined') {
ajax = new XMLHttpRequest();
}
return ajax;
}function saveUserInfo()
{
var msg = document.getElementById("msg");
var f = document.user_info;
var userName = f.user_name.value;
var userAge = f.user_age.value;
var userSex = f.user_sex.value;
var url = "/save_info.php";
var postStr = "user_name="+ userName +"&user_age="+ userAge +"&user_sex="+ userSex;
var ajax = InitAjax();
ajax.open("POST", url, true);
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
ajax.send(postStr);
ajax.onreadystatechange = function() {
if (ajax.readyState == 4 && ajax.status == 200) {
msg.innerHTML = ajax.responseText;
}
}
}
</script>
<div id="msg">
<form name="user_info">
姓名:<input type="text" name="user_name" /><br/>
年龄:<input type="text" name="user_age" /><br/>
性别:<input type="text" name="user_sex" /><br/>
<input type="button" value="提交表单" onClick="saveUserInfo()">
</form>save_info.php文件
<?PHP
$userage = $_POST["user_age"];
function writetofile($file_name,$data,$method="w")
{
$filenum=fopen($file_name,$method);
flock($filenum,LOCK_EX);
fputs($filenum,$data."\n");
fclose($filenum);
}
writetofile("yt.txt",$userage);
?>
目的就是通过txt文件有没有达到预期目标,
出现两个问题,
1.info.php运行了但是info_save.php没有运行了,为什么?
2.我试过单独执行info_save.php但是每次运行都是txt的第一行写入,怎么使它换行??
求高手帮忙,
function InitAjax()
{
var ajax=false;
try {
ajax = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
ajax = new ActiveXObject("Microsoft.XMLHTTP");
} catch (E) {
ajax = false;
}
}
if (!ajax && typeof XMLHttpRequest!='undefined') {
ajax = new XMLHttpRequest();
}
return ajax;
}function saveUserInfo()
{
var msg = document.getElementById("msg");
var f = document.user_info;
var userName = f.user_name.value;
var userAge = f.user_age.value;
var userSex = f.user_sex.value;
var url = "/save_info.php";
var postStr = "user_name="+ userName +"&user_age="+ userAge +"&user_sex="+ userSex;
var ajax = InitAjax();
ajax.open("POST", url, true);
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
ajax.send(postStr);
ajax.onreadystatechange = function() {
if (ajax.readyState == 4 && ajax.status == 200) {
msg.innerHTML = ajax.responseText;
}
}
}
</script>
<div id="msg">
<form name="user_info">
姓名:<input type="text" name="user_name" /><br/>
年龄:<input type="text" name="user_age" /><br/>
性别:<input type="text" name="user_sex" /><br/>
<input type="button" value="提交表单" onClick="saveUserInfo()">
</form>save_info.php文件
<?PHP
$userage = $_POST["user_age"];
function writetofile($file_name,$data,$method="w")
{
$filenum=fopen($file_name,$method);
flock($filenum,LOCK_EX);
fputs($filenum,$data."\n");
fclose($filenum);
}
writetofile("yt.txt",$userage);
?>
目的就是通过txt文件有没有达到预期目标,
出现两个问题,
1.info.php运行了但是info_save.php没有运行了,为什么?
2.我试过单独执行info_save.php但是每次运行都是txt的第一行写入,怎么使它换行??
求高手帮忙,
canshu=url+'&name='+document.getElementById('username').value;
$.ajax({
url: canshu,
type: 'GET',
async: false,
error: function(){
alert('Error loading XML document');
},
success: function(xml){
document.getElementById('username2_mes').innerHTML=xml;
}
});
}
<div id="msg"> </div>
div要有闭标记,否则会造成页面混乱2、save_info.php 中
要有输出的内容,你只是将提交的内容保存起来了
echo $userage = $_POST["user_age"]; 3、关于 每次运行都是txt的第一行写入,怎么使它换行??
writetofile("yt.txt","\r\n".$userage, "a+");