Question

我正忙着使用PHP脚本。

这就是我想要的： -人们可以输入其邮政编码。如果他们的邮政编码在table1中而不是..如果他们的邮政编码在table2中而不是...如果他们的邮政编码不在所有表中... -我可以插入带有其他数据示例名称的邮政编码。

但是我不知道如何构建它。我在互联网上进行搜索，试图获得所拥有的东西。但是，在其他表中进行搜索的功能却并非如此。它对我不起作用

这就是我所拥有的：

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Plusgas - Postcode invoeren</title>
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet"  href="bootstrap/bootstrap.min.css">
<link href="https://fonts.googleapis.com/css?family=Courgette" rel="stylesheet">
<script src="bootstrap/bootstrap.min.js"></script>
<style>
body{
    font-family: Courgette;
}
.submit{
  background-color: purple;
  color:white;
  text-size:24px;
  padding: 6px;
  border-radius: 5px;
  border:1px solid white;
  font-size: 24px;
}
.submit:hover{
  background-color: white;
  color: purple;
  box-shadow: 0px 0px 20px white;
}
h1{
	font-size: 14px;
}

td,th{
  padding: 4px;
  text-align: center;
}
</style>
</head>
<?php
$servername = "localhost";
$username="*";
$password="*";
$dbname="*";
$id="";
$postcode="";
$provincie="";
$website="";
$contactpersoon="";

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

//connect to mysql database
try{
	$conn =mysqli_connect($servername,$username,$password,$dbname);
}catch(MySQLi_Sql_Exception $ex){
	echo("error in connecting");
}
//get data from the form
function getData()
{
	$data = array();

	$data[1]=$_POST['postcode'];
	$data[2]=$_POST['provincie'];
	$data[3]=$_POST['website'];
	$data[4]=$_POST['contactpersoon'];

	return $data;
}
//search
if(isset($_POST['search']))
{
	$info = getData();
	$search_query="SELECT * FROM postcodes WHERE id = '$info[0]'";
	$search_result=mysqli_query($conn, $search_query);
		if($search_result)
		{
			if(mysqli_num_rows($search_result))
			{
				while($rows = mysqli_fetch_array($search_result))
				{
					$id = $rows['id'];
					$postcode = $rows['postcode'];
					$provincie = $rows['provincie'];
					$website = $rows['website'];
					$contactpersoon = $rows['contactpersoon'];
				}
			}else{
				echo("no data are available");
			}
		} else{
			echo("result error");
		}

}
//insert
if(isset($_POST['insert'])){
	$info = getData();
	$insert_query="INSERT INTO `postcodes`(`postcode`, `provincie`, `website`, `contactpersoon`) VALUES ('$info[1]','$info[2]','$info[3]','$info[4]')";
	try{
		$insert_result=mysqli_query($conn, $insert_query);
		if($insert_result)
		{
			if(mysqli_affected_rows($conn)>0){
				echo("Postcode is toegevoegd");

			}else{
				echo("Postcode is niet toegevoegd");
			}
		}
	}catch(Exception $ex){
		echo("error inserted".$ex->getMessage());
	}
}

?>
<body>
	<div class="container-fluid">
	  <div class="row">
	    <div class="col-lg-4">

<form method ="post"   action="">
	<h1>ID nummer (voor filterern)</h1>
  <input type="number" name="id"  class="form-control" placeholder="ID No. /Automatic Number Genrates" value="<?php echo($id);?>" disabled>
	<div class="form-group row">
	<div class="col-xs-6">
		<h1>Postcode</h1>
	<input type="text" name="postcode" class="form-control" placeholder="Postcode" value="<?php echo($postcode);?>" required>
</div>
<div class="col-xs-6">
	<h1>Provinicie</h1>
	<input type="text" name="provincie" class="form-control" placeholder="Provinicie" value="<?php echo($provincie);?>" required>
</div>
</div>
		<h1>Website</h1>
  <select name="website" class="form-control" value="<?php echo($website);?>">
    <option value="websiteZH">Website Zuid-Holland</option>
    <option value="websiteNH">Website Noord-Holland</option>
    <option value="websiteZL">Website Zeeland</option>
    <option value="websiteUT">Website Utrecht</option>
  </select>
</div>

<div class="col-xs-6">
	<h1>Contactpersoon</h1>
  	<input type="text" name="contactpersoon" class="form-control" placeholder="Contactpersoon" value="<?php echo($contactpersoon);?>" required>
</div>
</div>

	<div>
		<input type="submit" class="btn btn-success btn-block btn-lg" name="insert" value="Add">
	</div>
</form>
</div>

    <div class="col-lg-8">
			<h2>Student Data</h2>
<?php
$servername = "localhost";
$username = "*";
$password = "*";
$dbname = "*";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT id,postcode,provincie,website,contactpersoon FROM postcodes";
$result = $conn->query($sql);

echo "<table border='1'>
<tr>
<th>Search ID</th>
<th>Postcode</th>
<th>Provinicie</th>
<th>Website</th>
<th>Contactpersoon</th>

</tr>";

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {

      echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['postcode'] . "</td>";
echo "<td>" . $row['provincie'] . "</td>";
echo "<td>" . $row['website'] . "</td>";
echo "<td>" . $row['contactpersoon'] . "</td>";

echo "</tr>";


    }
} else {
    echo "0 results";
}

$conn->close();
?>
</div>
</div>
</body>
</html>

Answer 1

我不确定要实现什么，出了什么问题，但是我可以肯定地看到一个问题-您的函数getData（）返回索引为1..4的数组。然后在搜索部分得到它，然后搜索$ info [0]，并且永远不会设置它。

第二个-这是危险的编码方式。如果有人要使用您的代码，则可以使用SQL注入。不用说，即使有人迷惑，也可以使用'（撇号）并崩溃SQL查询。您应该始终转义传递给查询的数据，即使用real_escape_string（）或准备好的语句。

连接表PHP

1 个答案: