Php jquery ajax user logon instance code

Source: Internet
Author: User
Tags require

Php Jquery does not require any new login, so it is easy to write your own
Added a check to determine whether the logon status is displayed after refresh if it is logged on.
Logon page File
Html page

The code is as follows: Copy code

<Script type = "text/javascript" src = "jquery-1.4.3.js" mce_src = "jquery-1.4.3.js"> </script>

<Script type = "text/javascript">

$ (Document). ready (function () {// onload event handler function of DOM
$ ("# Button"). click (function () {// processing function when a button is clicked
Postdata (); // execute the postdata function when the button is clicked.
});

});

Function postdata () {// submit data function
$. Ajax ({// call jquery's ajax method
Type: "POST", // Set the ajax method to submit data
Url: "login_ OK .php", // submit the data to OK. php.
Data: "writer =" + $ ("# writer "). val () + "& pass =" + $ ("# pass "). val (), // input the value in the writer box as the submitted data
Success: function (msg) {// callback after successful submission. The msg variable is output by OK. php.
$ ("# Div2" cmd.html (msg); // if necessary, you can display the value of the msg variable to a DIV element.
}
});
}


</Script>
<Script type = "text/javascript">
<! --
Function init (){
$. Ajax ({
Type: "GET ",
Url: "if_login.php ",
Data: "ts =" + new Date (). getTime (),
Success: function (msg) {// callback after successful submission. The msg variable is output by OK. php.
If (msg = 'true '){
$ ("# Div2" pai.html ("login OK ");
}
}
});
}
// -->
</Script>
<Body onLoad = "init ();">
<Div id = "div2">
<Input name = "writer" id = "writer" type = "text" value = ""/>
<Input name = "pass" id = "pass" type = "password" value = ""/>
<Input type = "submit" name = "button" id = "button" value = "submit"/>
</Div>
</Body>


Send to file

The code is as follows: Copy code

<? Php
Session_start ();
Require "conn. php ";
$ Username = $ _ POST ['writer'];
$ Password = $ _ POST ['pass'];
Mysql_select_db ($ database_lr, $ lr );
$ SQL = "SELECT * FROM admin WHERE username = '$ username' AND password =' $ password '";
$ Result = mysql_query ($ SQL );
If (mysql_num_rows ($ result)> 0)
{
// Login sucess
$ _ SESSION ['login _ admin'] = "Admin ";
Echo "login OK ";

}
Else
{
// The user ID found
Echo "Login failed, ID error or expired .";
}

?>

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