我创建了一个表单，当输入框的内容更改并提交时，应该更新输入框中的表格内容，数据库应该更新。 这是我的第一页:
<body>
<form action="qa1.php" method="post">
<ul>
<li><a class="active" href="df1.php">Disease</a></li>
<li><a href="drug.php" >Drug</a></li>
<li><a href="#about">Interaction</a></li>
<a href="#" class="dropbtn">Alternate Drug</a>
</ul>
<?php
$query = "SELECT * FROM disease;";
$result = mysqli_query($dp, $query);
echo "<table border=5>
<tr>
<th>Select</th>
<th>Edit</th>
<th>Disease ID</th>
<th>Disease</th>
<th>Sub Disease</th>
<th>Associated Disease</th>
<th>Ethinicity</th>
<th>Source</th>
</tr>";
while($row = mysqli_fetch_assoc($result)) {
echo "<tr>";
echo "<td><input type='radio' name='select' value=".$row['Disease_id']."/> </td>";
echo "<td>".$row{'Disease_id'}."</td>";
echo "<td><a href='edit.php?Disease_id=".$row['Disease_id']."'>edit</a></td>";
echo "<td>".$row{'Disease'}."</td>";
echo "<td>".$row{'SubDisease'}."</td>";
echo "<td>".$row{'Associated_Disease'}."</td>";
echo "<td>".$row{'Ethinicity'}."</td>";
echo "<td>".$row{'Source'}."</td>";
echo "</tr>";}
echo "</table>";
$selectedRow=$_POST['select'];
?>
<div>
<table border="0" align="center" style="border-spacing: 40px 30px;">
<caption><strong>QualityAnalysis:</br></br></strong></caption></br></br>
<TABLE BORDER="0" CELLSPACING="0" CELLPADDING="4" WIDTH="40%">
</br><div><center>
<button style="color: red"><a href="adddf1.php" target="_self" name="Add" value="Add">Add</a></button>
<input type = 'submit' value = 'Delete' name = 'submitdelete' button style="color: red"></button>
<input type = 'submit' value = 'Update' name = 'submitupdate'>
</center></div> </TABLE>
</body>
</html>
这是我的编辑页面:edit.php
<body>
<form action="edit.php" method="post">
<ul>
<li><a class="active" href="df1.php">Disease</a></li>
<li><a href="drug.php" >Drug</a></li>
<li><a href="#about">Interaction</a></li>
<a href="#" class="dropbtn">Alternate Drug</a>
</ul>
<div>
<?php
$conn = mysqli_connect('localhost','root','','tool');
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_error());
}
if(isset($_GET['Disease_id']))
{
if(isset($_POST['Update']))
{
$id=$_GET['Disease_id'];
$name=$_POST['Ethinicity'];
$query3=mysqli_query("update disease set Ethinicity='$name', where Disease_id='$id'");
if($query3)
{
header('location:qa1.php');
}
}
$query1=mysqli_query("select * from disease where Disease_id='$id'");
$query2=mysqli_fetch_array($query1);
?>
<table border="2" align="center" style="border-spacing: 40px 30px;">
<caption><strong>Edit</br></br></strong></caption></br></br>
<tr>
<td>Ethinicity<input type="text" list="Ethinicity" name="Ethinicity" value="<?php echo $row['Ethinicity'];?>"/>
<input type="hidden" name="Disease_id" value="<?php echo $row['Disease_id']; ?>"/>
<input type="submit" name="Update" value ="Update">
<?php
}
?>
</center></div></div>
</form>
但是数据库中的数据没有更新。谁能帮我解决这个问题吗？
最佳答案
首先，mysqli_query需要连接参数。
其次，您的 UPDATE 查询存在语法错误，不需要逗号。
所以，应该是:
$query3=mysqli_query($conn, "update disease set Ethinicity='$name' where Disease_id='$id'");
$query1=mysqli_query($conn, "select * from disease where Disease_id='$id'");
此外，if($query3) 不会检查该行是否已成功插入，请使用 mysqli_affected_rows相反。
关于php - 使用php以html形式更新mysql，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37359368/