我正在制作一个Web应用程序，而且每次重新加载页面时，我都无法弄清楚如何阻止PHP对我的数据库进行查询。
基本上，当提交表单时，我希望PHP对添加变量的数据库进行查询，但是，因为它已将这些值保存在$ _POST中，所以它每隔时间进行查询我用相同的值刷新。
我希望它能够进行查询，然后可能取消设置$ _POST值或某些内容以使其不满足if条件，并且当提交按钮没有时，它会阻止它使用重复值查询数据库用新值更新它们。
对不起，如果那令人费解，我试图尽我所能解释我的问题。
这是PHP
<?php
//Require login gile -- Fatal error if not found
require_once('login.php');
require_once('app.html');
//Connect w/ MySQL database [login variables]
$con = new mysqli($hn, $un, $pw, $db);
//If connection error -> kill application and display error
if ($con->connect_error) die('Connect Error (' . $con->connect_errno . ') '. $con->connect_error);
//If both goal and difficulty are set
if (!empty($_POST['goal']) && !empty($_POST['difficulty'])) {
$goal = get_post($con, 'goal');
$difficulty = get_post($con, 'difficulty');
$query = "INSERT INTO items VALUES" . "('$goal', 1, '$difficulty', NULL)";
$result = $con->query($query);
if(!$result) echo "INSERT failed: $query<br>" . $con->error . "<br><br>";
}
unset($_POST['goal']);
unset($_POST['difficulty']);
//close connection
$con->close();
function get_post($conn, $var) {
return $conn->real_escape_string($_POST[$var]);
}
?>
HTML
<!DOCTYPE html>
<html>
<head>
<title>Testing</title>
<!--Page Style-->
<link type="text/css" rel="stylesheet" href="Style.css">
<!--jQuery-->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>
<!--RateYo | http://prrashi.github.io/rateYo/-->
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/rateYo/2.0.1/jquery.rateyo.min.css">
<script src="https://cdnjs.cloudflare.com/ajax/libs/rateYo/2.0.1/jquery.rateyo.min.js"></script>
</head>
<body>
<div id="parent1">
<form id="goalForm" method="post" action="PHP.php">
<!--Goal Value-->
<input id="goalName" type="text" name="goal" maxlength="150" placeholder="Type your goal here!">
<!--Star Rating-->
<div id="rateYo"></div><br>
<!--Value from stars to be submitted-->
<input id="dif" type="hidden" name="difficulty">
<input class="but1" id="firstSubmit" type="button" value="Submit">
<input class="but1" id="submitton" type="submit" style="display:none" value="Difficulty?">
</form>
</div>
<script>
$("#goalName").val("");
//First Submit Onclick
$("#firstSubmit").click(function() {
if($("#goalName").val() == "") {
alert("Please insert a value.");
}
else {
$("#goalName").fadeOut("slow" , function() {
$("#rateYo").rateYo({
numStars: 10,
rating: 0,
fullStar: true,
starWidth: "70px",
ratedFill: "#E74C3C",
maxValue: 10
});
$("#rateYo").fadeIn("slow");
});
$(this).hide();
$("#submitton").show();
}
});
//Star Submit
$("#goalForm").submit(function() {
var $rateYo = $("#rateYo").rateYo();
var rating = $rateYo.rateYo("rating");
if($("#goalName").val() == "") {
alert("Please insert a value.");
return false;
}
else if(rating == 0) {
alert("Please set a difficulty level.");
return false;
}
else {
$("#dif").val(rating);
}
});
</script>
答案 0 :(得分：2)
试试吧PLZ
<?php
//Require login gile -- Fatal error if not found
require_once('login.php');
require_once('app.html');
//Connect w/ MySQL database [login variables]
$con = new mysqli($hn, $un, $pw, $db);
//If connection error -> kill application and display error
if ($con->connect_error) die('Connect Error (' . $con->connect_errno . ') '. $con->connect_error);
//If both goal and difficulty are set
if (!empty($_POST['goal']) && !empty($_POST['difficulty'])) {
$goal = get_post($con, 'goal');
$difficulty = get_post($con, 'difficulty');
$query = "INSERT INTO items VALUES" . "('$goal', 1, '$difficulty', NULL)";
$result = $con->query($query);
if(!$result) echo "INSERT failed: $query<br>" . $con->error . "<br><br>";
//You can write samepage name in location as window.location='abc.php'
echo "<script>window.location=''</script>";
}
//close connection
$con->close();
function get_post($conn, $var) {
return $conn->real_escape_string($_POST[$var]);
}
?>