Php 更新数据库更新
我在更新数据库时遇到问题。我的sql update语句是否有问题？我检查了sql语句，它说数据库中没有记录。我不知道该怎么办Php 更新数据库更新,php,html,sql,Php,Html,Sql,我在更新数据库时遇到问题。我的sql update语句是否有问题？我检查了sql语句，它说数据库中没有记录。我不知道该怎么办 <!-- template for mySql database access. --> <!DOCTYPE html> <html> <head> <title>CRUD</title> <link href="/sandv
<!-- template for mySql database access. -->
<!DOCTYPE html>
<html>
<head>
<title>CRUD</title>
<link href="/sandvig/mis314/assignments/style.css" rel="stylesheet" type="text/css">
</head>
<div class="pageContainer centerText">
<h3>CRUD (Create, Read, Update, & Delete) Database</h3>
<?php
//include database connection
include("DatabaseConnection2.php");
//connect to database
$link = fConnectToDatabase();
//Retrieve parameters from querystring and sanitize
$nameF = fCleanString($link, $_GET['nameF'], 15);
$nameL = fCleanString($link, $_GET['nameL'], 15);
$deleteID = fCleanNumber($_GET['deleteID']);
$updateID = fCleanNumber($_GET['updateID']);
$updateID2 = fCleanNumber($_GET['updateID2']);
//Populate Textbox
if (!empty($updateID)) {
$sql = "SELECT NameL, NameF
FROM customertbl
WHERE custID = '$updateID'";
mysqli_query($link, $sql) or die('Delete error: ' . mysqli_error($link));
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
$row = mysqli_fetch_array($result);
$strFName2 = $row[NameF];
$strLName2= $row[NameL];
}
?>
<hr>
<form class="formLayout">
<div class="formGroup">
<label>First name:</label>
<input name="nameF" type="text" autofocus value="<? echo $strFName2; ?>">
</div>
<div class="formGroup">
<label>Last name:</label>
<input name="nameL" type="text" value="<? echo $strLName2; ?>">
</div>
<div class="formGroup">
<label> </label>
<button>Submit</button>
<input type="hidden" name="updateID2" value="<? echo $updateID; ?>">
</div>
</form>
<?php
//Update
if (!empty($updateID2))
{
$sql = "UPDATE customertbl
SET NameL = '$strFName2', NameF ='$strLName2'
WHERE custID = '$updateID2' ";
mysqli_query($link, $sql) or die('Insert error: ' . mysqli_error($link));
}
//Insert
if (!empty($nameF) && !empty($nameL)) {
$sql = "Insert into customertbl (NameL, NameF)
VALUES ('$nameL', '$nameF')";
mysqli_query($link, $sql) or die('Insert error: ' . mysqli_error($link));
}
//Delete
if (!empty($deleteID)) {
$sql = "Delete from customertbl WHERE CustID= '$deleteID' ";
mysqli_query($link, $sql) or die('Delete error: ' . mysqli_error($link));
}
//List records
$sql = 'SELECT custID, NameF, NameL
FROM customertbl order by custID';
//$result is an array containing query results
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
echo "<p>" . mysqli_num_rows($result) . " records in the database</p>";
?>
<table class="simpleTable">
<tr>
<th>Cust. ID</th>
<th>F. Name</th>
<th>L. Name</th>
<th>Delete</th>
<th>Update</th>
</tr>
<?php
// iterate through the retrieved records
while ($row = mysqli_fetch_array($result)) {
//Field names are case sensitive and must match
//the case used in sql statement
$custID = $row['custID'];
echo "<tr>
<td>$custID</td>
<td>$row[NameF]</td>
<td>$row[NameL]</td>
<td><a href='?deleteID=$custID'>Delete</a></td>
<td><a href='?updateID=$custID'>Update</a></td>
</tr>";
}
?>
</table>
</div>
</body>
</html>
积垢
CRUD（创建、读取、更新和删除）数据库
有问题的代码块
//Update
if (!empty($updateID2))
{
$sql = "UPDATE customertbl
SET NameL = '$strFName2', NameF ='$strLName2'
WHERE custID = '$updateID2' ";
mysqli_query($link, $sql) or die('Insert error: ' . mysqli_error($link));
}
引用变量$strFName2
和$strLName2
，这两个变量是仅按条件填充的变量
//Populate Textbox
if (!empty($updateID)) {
$sql = "SELECT NameL, NameF
FROM customertbl
WHERE custID = '$updateID'";
mysqli_query($link, $sql) or die('Delete error: ' . mysqli_error($link));
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
$row = mysqli_fetch_array($result);
$strFName2 = $row[NameF];
$strLName2= $row[NameL];
}
由于变量$strFName2
和$strLName2
在更新
SQL查询期间未定义，因此您无法看到所需的结果
查询应该引用$nameF
和$nameL
，因为这些变量总是定义的（不包含在条件中），表单输入在名称属性中使用nameF
和nameL
$sql = "UPDATE customertbl
SET NameL = '$nameF', NameF ='$nameL'
WHERE custID = '$updateID2';";
您还需要修复DELETE
查询以引用列custID
，而不是custID
，因为您的模式似乎使用了前者
$sql = "Delete from customertbl WHERE custID= '$deleteID' ";
请告诉我们您得到的错误以及它属于代码的哪一行。这让社区更容易帮助你。我没有犯错误，我只是没有得到我想要的结果。当我单击更新超链接时，它会填充文本框。您应该能够编辑文本框，然后单击提交。它更新了数据库字段，解决了这些问题。但是我仍然有一个问题。每次我更新时，它都会将其分配给一个新id。它不会更新我想要的id字段to@TimothyWong这是因为INSERT
代码块将始终运行，因为它不包含在限制它仅在$updateID2
为空时运行的条件中。