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php - 使用php以html形式更新mysql

转载 作者:行者123 更新时间:2023-11-29 20:59:31 25 4
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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']."/>&nbsp;</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'");
<小时/>

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/

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