我正在尝试首先获取ID来查询数据库并将结果打印在表中(此部分有效)。然后,我想获取用户提供的ID,并使用它使用PHP更新数据库中的信息。我想使用第二种形式上的输入作为更新数据库的值。要更改的表是customers
,它具有字段ID,NAME,ADDRESS。我不希望用户能够更改ID。
表格1:
<form method="post" action="">
<p style="margin-top: 70px;">Please type the ID of the person you wish to add to change their data</p>
<p style="margin-bottom: 0px;">ID</p>
<input style="color:black" type="text" name="id" placeholder="10001">
<input style="color:lightblue;background-color: rgb(80,80,80);margin-top: 7px; " type="submit" value="Submit">
</form>
表格2:
<form method="post" action="">
<p>New Information for Customer with ID entered above</p>
<input style='color:black;' type='text' name='newName' placeholder='Name Change'>
<input style="color:black;" type="text" name="newAddress" placeholder="New Address">
<input style="color:lightblue;background-color: rgb(80,80,80);margin-top: 7px; " type="submit" name="submitForm2" value="Submit">
</form>
这是我当前要求的php,但它不起作用,检查值是否设置的$ _POST返回false。
<?php
session_start();
if (isset($_POST["id"])){
$servername = 'localhost';
$user = 'root';
$pass = '';
$db = 'the_sports_store';
$conn = new mysqli($servername,$user, $pass, $db);
// Check connection
if ($conn->connect_error) {
echo '<script language="javascript">';
echo 'alert("DB Connection Failed:")';
echo '</script>';
die("" . $conn->connect_error);
}
$sessionID = $_SESSION["ID"];
$newName = $_SESSION["newName"];
$newAddress = $_SESSION["newAddress"];
var_dump($newName);
$sql = "SELECT * FROM `customers` WHERE ID='$sessionID';";
//display the current record, allow user input to alter it, then display new data
if ($conn->query($sql) == TRUE) {
echo"<div class='col-10'>";
echo"<table>";
echo"<tr>
<td align='justify'><b>ID</b></td>
<td align='justify'><b>NAME</b></td>
<td align='justify'><b>ADDRESS</b></td>
</tr>";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
echo "<tr><td style='padding: 10px;'>{$row['ID']}</td><td>{$row['NAME']}</td><td>{$row['ADDRESS']}</td></tr>";
echo "</table>";
echo "</div>";
if(!empty($_POST["newName"]) && !empty($_POST["newAddress"])){
echo '<script language="javascript">';
echo 'alert(',$sessionID,');';
echo '</script>';
$newName = $_POST["newName"];
$newAddress = $_POST["newAddress"];
$sqlChange = "UPDATE `customers`
SET `NAME` = '$newName', `ADDRESS` = '$newAddress'
WHERE `ID` = '$sessionID';";
if ($conn->query($sqlChange) === TRUE) {
echo '<script language="javascript">';
echo 'alert("Update Successful.")';
echo '</script>';
} else {
echo '<script language="javascript">';
echo 'alert("Error. Update Unsucessful.")';
echo '</script>';
}
}else if(!empty($_POST["newName"])){
$newName = $_POST["newName"];
$sqlChange = "UPDATE `customers` SET `NAME` = '$newName' WHERE `ID` = '$sessionID'";
echo '<script language="javascript">';
echo 'alert(',$newName,');';
echo '</script>';
if ($conn->query($sqlChange) === TRUE) {
echo '<script language="javascript">';
echo 'alert("Update Successful.")';
echo '</script>';
} else {
echo '<script language="javascript">';
echo 'alert("Error. Update Unsucessful.")';
echo '</script>';
}
}else if(!empty($_POST["newAddress"])){
$newName = $_POST["newAddress"];
$sqlChange = "UPDATE `customers` SET `ADDRESS` = '$newAddress' WHERE `ID` = '$sessionID'";
echo '<script language="javascript">';
echo 'alert(',$sessionID,');';
echo '</script>';
if ($conn->query($sqlChange) === TRUE) {
echo '<script language="javascript">';
echo 'alert("Update Successful.")';
echo '</script>';
} else {
echo '<script language="javascript">';
echo 'alert("Error. Update Unsucessful.")';
echo '</script>';
}
} else{
echo '<script language="javascript">';
echo 'alert(',$sessionID,');';
echo '</script>';
}
}
$conn->close();
}
?>
您的问题是您的PHP代码仅在设置了id的情况下才会执行。因此,当您发布第二个表单时,代码将永远不会执行。
将此if(!empty($_POST["newName"]) && !empty($_POST["newAddress"]))
和所有elseif / else移至初始if之外。
另外,我有义务向您介绍SQL注入以及如何避免它:如何防止PHP中的SQL注入?
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