我的HTML带有文本框和选择框。我希望用户从下拉选择框中选择一个固定号码,该号码从我的数据库中选取号码。然后,我希望用户在文本框中输入新的月租金。一旦用户点击提交按钮,我希望它根据已选择的单位数更新数据库月租。我可以从数据库中获取统一号码,但似乎无法从文本框中获取信息并在数据库中进行更改。
这是我的代码
Changeback.php(后端)
<?php
include "connect.php";
if(isset($_POST['MonthlyRent'])){
$MonthlyRent = $_POST['MonthlyRent'];
$FlatCode = $_POST['FlatCode'];
$query = "UPDATE MonthlyRent set MonthlyRent = ($MonthlyRent) where FlatCode = ($FlatCode) ";
mysqli_query($con, $query) or die ("Invalid query");
$numrows = mysqli_affected_rows($con);
echo "number of affected rows is " . $numrows;
exit( print_r( $_POST ) );
mysqli_close($con);
}
?>
&#13;
change.php(前端)
<html><head><title>Connect to Database</title></head><body>
<font size="4"> Enter owner details</font><br><br>
<form action="changeback.php" method="post" >
Monthly Rent <input type="text" name="MonthlyRent">
Flat Code <select name="FlatCode">
<?php
$con = mysqli_connect("localhost", "root", "root") or die ("No connection");
mysqli_select_db($con , "flat") or die ("db will not open");
$query = "SELECT distinct FlatCode from FLAT";
$result = mysqli_query($con, $query) or die("Invalid query");
while($rows = mysqli_fetch_array($result)) {
echo "<option value=\"" . $rows[0] . "\">" . $rows[0] . "</option>";
}
echo "</select>";
mysqli_close($con);
?>
<input type="submit" value="Submit Value">
</form></body></html>
&#13;
答案 0 :(得分:0)
问题出在您的UPDATE语句中,您(1)已将MonthlyRent
列名放在FLAT
表名称所在的位置,以及(2)未在单位周围放置引号代码值。
变化:
$query = "UPDATE MonthlyRent set MonthlyRent = ($MonthlyRent) where FlatCode = ($FlatCode) ";
要:
// Always escape user input to reduce risk of errors and security holes
$escapedMonthlyRent = mysqli_real_escape_string($con, $MonthlyRent);
$escapedFlatCode = mysqli_real_escape_string($con, $FlatCode);
$query = "update FLAT set MonthlyRent = '$escapedMonthlyRent' "
."where FlatCode = '$escapedFlatCode'";
答案 1 :(得分:0)
这不是一个答案,只是一个长篇评论。
如果您有时间(几个小时),我强烈建议您前往PHPAcademy并进行一些免费教程。
这是一个开始:
https://phpacademy.org/videos/php-and-mysql-with-mysqli
如果您希望人们登录您的网站,则必须提供Register & Login
教程。
此外,新波士顿还有一些很棒的(免费)教程,你可以查看。