[英]Checking if Username exists not working
我認為問題與if(data == 1)
有關,此刻,警報框一直說可用,而當我將其更改為===
,警報框一直說不可用。 我不確定問題出在哪里。
這是我的register.html
<form name="registerform" id="registerform" action="register.php" method="post">
<div class="form_settings">
<p>
<label for="firstname">First Name:</label>
<input name="firstname" id="firstname" maxlength="32" type="text" placeholder="Optional" />
</p>
<p>
<label for="username" id="usernameText">Username:</label>
<input data-parsley-pattern="^[A-Za-z0-9_]{3,15}$" data-parsley-length="[3, 15]" name="username" id="username" maxlength="32" type="text" data-parsley-error-message="Username needs to be between 3 and 15 characters. Case sensitive. No special characters allowed." required/>
</p>
<p>
<label for="password1">Password:</label>
<input name="password1" id="password1" data-parsley-pattern="^[A-Za-z0-9_-]{5,25}$" data-parsley-length="[5, 25]" type="password" data-parsley-equalto="#password2" data-parsley-error-message="Passwords must match. Needs to be between 5 and 25 characters. Case sensitive. No special characters allowed." required/>
</p>
<p>
<label for="password2">Confirm Your Password:</label>
<input name="password2" id="password2" data-parsley-length="[5, 25]" data-parsley-error-message="Passwords must match. Needs to be between 5 and 25 characters. Case sensitive. No special characters allowed." data-parsley-pattern="^[A-Za-z0-9_-]{5,25}$" type="password" data-parsley-equalto="#password1" required/>
</p>
<p>
<label for="email">E-Mail:</label>
<input data-parsley-trigger="change" name="email" id="email" maxlength="1024" type="email" required/>
</p>
<p style="padding-top: 15px"><span> </span>
<input type="submit" class="submit" name="createacc" value="Register" />
</p>
</div>
</form>
這是我的javascript
$(document).ready(function () {
$('input[name="createacc"]').click(function (e) {
var username = $('input[name="username"]').val();
if (username != '') {
e.preventDefault();
$.ajax({
type: 'POST',
data: {
username: username
},
url: 'checkuser.php',
success: function (data) { //Receives the data from the php code
if(data == 1){
//show that the username is available
alert("available");
}else{
//show that the username is NOT available
alert("Unavailable");
}
},
error: function (xhr, err) {
console.log("readyState: " + xhr.readyState + "\nstatus: " + xhr.status);
console.log("responseText: " + xhr.responseText);
}
});
$('#registerform').parsley();
}
});
});
這是我的checkuser.php
<?php
define('DB_SERVER', "localhost");
define('DB_USER', "XXX");
define('DB_PASSWORD', "XXX");
define('DB_DATABASE', "XXX");
$mysqli = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_DATABASE);
if(mysqli_connect_errno()) {
echo "Connection Failed.";
exit();
}
$username = mysqli_real_escape_string($_POST['username']);
$result = mysqli_query("SELECT `username` FROM `accounts` WHERE username = '$username'");
if(mysqli_num_rows($result)>0){
echo json_encode(0);
}else{
echo json_encode(1);
}
?>
更新我已經嘗試了這篇文章的評論部分中發布的所有建議,但到目前為止沒有任何效果。 不知道該怎么辦= /
您實際上在返回的字符串中有錯誤。
2件事情: mysqli_real_escape_string() expects exactly 2 parameters
而mysqli_query() expects 2 parameters
。
現在這兩種方法都需要2個參數,並且由於您只提供了1個參數,因此它返回錯誤字符串,而不是0
和1
。 因此,將您的mysqli連接變量添加為這兩種方法中的第一個參數。 這是替換代碼。
$username = mysqli_real_escape_string($mysqli, $_POST['username']);
$result = mysqli_query($mysqli,"SELECT `username` FROM `ncs_users` WHERE username = '$username'");
並出於安全原因,嘗試使用參數化查詢而不是像在查詢中對$username
那樣進行串聯。
你可以回來
if(mysqli_num_rows($result)>0){
echo json_encode(array('vaild' => false));
die;
}
else{
echo json_encode(array('vaild' => true));
}
在javascript中設置:
$.ajax({
type: 'POST',
dataType: "json",
data: {
username: username
},
url: 'checkuser.php',
success: function (data) { //Receives the data from the php code
if(data.valid == true){
//show that the username is available
alert("available");
}else{
//show that the username is NOT available
alert("Unavailable");
}
},
error: function (xhr, err) {
console.log("readyState: " + xhr.readyState + "\nstatus: " + xhr.status);
console.log("responseText: " + xhr.responseText);
}
});
$('#registerform').parsley();
}
});
我希望這能幫到您。
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