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php - 如何在 PHP 中添加一个简单的验证?

转载 作者:行者123 更新时间:2023-11-28 15:18:42 25 4
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我如何添加一个简单的验证来检查每个输入是否不为空?

我只是针对一个简单的验证,如果至少有一个表单为空,它会在 php 文件中显示错误。并在所有内容完全填满后继续将输入添加到数据库中。我已经为 html 表单的表单字段中的每个输入设置了变量。

HTML代码:

<html>
<head>
<title>FEATHER FRIENDS PIZZA SHOP</title>
</head>
<img src="pics/logo.png">
<style>
img {
display: block;
margin: auto;
width:100%;
}

input[type=text], select {
width: 100%;
padding: 12px 20px;
margin: 8px 0;
display: inline-block;
border: 1px solid #ccc;
border-radius: 4px;
box-sizing: border-box;
}

input[type=submit] {
width: 100%;
background-color: #4CAF50;
color: white;
padding: 14px 20px;
margin: 8px 0;
border: none;
border-radius: 4px;
cursor: pointer;
}

input[type=reset] {
width: 100%;
background-color: #bfac2c;
color: white;
padding: 14px 20px;
margin: 8px 0;
border: none;
border-radius: 4px;
cursor: pointer;
}

input[type=reset]:hover {
background-color: #c43848;
}

input[type=submit]:hover {
background-color: #c43848;
}

div {
background-image: url('pics/bg.png');
background-attachment: fixed;
background-repeat: no-repeat;
background-size: 100% 100%;
padding: 100px;
}
body {
display: block;
margin: auto;
width: 100%;
height: 100%;
background-color: #f6f6d4;
background-attachment: fixed;
background-repeat: no-repeat;
background-size: 100% 100%;
}
</style>
<div>
<body>
<img src="pics/pizza.png">
<form action="http://localhost/insert.php" method="post">

<img src="pics/name.png">
<input type="text" id="fname" name="name" placeholder="Your full name...">

<img src="pics/size.png">
<select id="size" name="size">
<option value="small">Small</option>
<option value="medium">Medium</option>
<option value="large">Large</option>
<option value="extra large">Extra Large</option>
</select>

<img src="pics/crust.png">
<select id="crust" name="crust">
<option value="pan">Pan</option>
<option value="thin">Thin</option>
<option value="stuffed">Stuffed</option>
<option value="handtossed">Hand-Tossed</option>
<option value="deepdish">Deep Dish</option>
</select>

<img src="pics/garnish.png">
<input type="text" id="garnish" name="garnish" placeholder="Write your choices here! Ex: Pepperoni, Cheese, Bacon, Mushroom">

<img src="pics/address.png">
<input type="text" id="address" name="address" placeholder="Where should we deliver?">

<img src="pics/contact.png">
<input type="text" id="contact" name="contact" placeholder="What is your contact number?">

<input type="submit" value="Submit">
<input type="reset" value="Reset your Order?">
</form>
</div>
</body>
</html>

PHP代码:

<?php

$link = mysqli_connect("localhost", "root", "", "pizza");

if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}

$name = mysqli_real_escape_string($link, $_POST['name']);
$size = mysqli_real_escape_string($link, $_POST['size']);
$crust = mysqli_real_escape_string($link, $_POST['crust']);
$garnish = mysqli_real_escape_string($link, $_POST['garnish']);
$address = mysqli_real_escape_string($link, $_POST['address']);
$contact = mysqli_real_escape_string($link, $_POST['contact']);

$sql = "INSERT INTO deliver (name, size, crust, garnish, address, contact)
VALUES ('$name', '$size', '$crust', '$garnish', '$address', '$contact')";
if(mysqli_query($link, $sql)){
echo "Data successfully Saved.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

mysqli_close($link);
?>

最佳答案

下面是一行简单的代码,用于检查是否填充了所有表单字段:

<?php

$link = mysqli_connect("localhost", "root", "", "pizza");

if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}

$name = mysqli_real_escape_string($link, $_POST['name']);
$size = mysqli_real_escape_string($link, $_POST['size']);
$crust = mysqli_real_escape_string($link, $_POST['crust']);
$garnish = mysqli_real_escape_string($link, $_POST['garnish']);
$address = mysqli_real_escape_string($link, $_POST['address']);
$contact = mysqli_real_escape_string($link, $_POST['contact']);


/*check if all the fields are not empty*/
if( $name != "" && $size != "" && $crust != "" && $garnish != "" && $address != "" && $contact != "") {

$sql = "INSERT INTO deliver (name, size, crust, garnish, address, contact) VALUES ('$name', '$size', '$crust', '$garnish', '$address', '$contact')";

if(mysqli_query($link, $sql)){
echo "Data successfully Saved.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

} else {
echo "Form incomplete";
}

mysqli_close($link);

?>

关于php - 如何在 PHP 中添加一个简单的验证?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46628580/

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