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php - 在php和mysql中将数据插入数据库

转载 作者:太空宇宙 更新时间:2023-11-03 10:41:51 25 4
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我已尝试使用以下代码将数据插入数据库。正在连接数据库。但是数据没有被添加到数据库中。我不知道我哪里出错了。谁能帮我解决这个问题?

    <html>
<head>

<title>registration</title>
<meta charset="UTF-8">
<link href="site.css" rel="stylesheet">
<div align="center">
<link rel="stylesheet" href="mine.css"/>

<table border="0" align="center" style="border-spacing: 40px 20px;">

<align="center"> <td>
</head>

<body bgcolor=" #b3ffe0">

<style>

html {
font-family: "Lucida Sans", sans-serif;
}
ul li {display: block;position: relative;float: left;border:1px }
ul li a {display: block;text-decoration: none; white-space: nowrap;color:#fff;}
ul {
list-style-type: none;
padding: 2px ;
margin-left: auto;
background-color: #666;
}
li a, .dropbtn {
display: inline-block;
color: white;
text-align: center;
padding: 10px 20px;
text-decoration: none;

}
li a:hover, .dropdown:hover .dropbtn {
background-color: #111;
}
</style>
</head>
<body>
<form action="df1.php" method="post">
<ul>
<li><a class="active" href="df.html">Disease</a></li>
<li><a href="drug.html" align="_self">Drug</a></li>
<li><a href="#about">Interaction</a></li>
<a href="#" class="dropbtn">Alternate Drug</a>
</ul>


<div>


<table border="2" align="center" style="border-spacing: 40px 30px;">
<caption><strong>DISEASE DETAILS:</br></br></strong></caption></br></br>
<tr>
<td><center> Disease_ID
<select name="Day">
<option value="1">1</option>
<option value="2">2</option>
</select></center>
</td>
<td>Disease
<select name="DisType">
<option value="Select">Select</option>
<option value="Acute">Acute</option>
<option value="Chronic">Chronic</option>
<option value="Acquired">Acquired</option>
</select>
</td>

<td>SubDisease
<select name="DisType">
<option value="Select">Select</option>
<option value="Acute">Acute</option>
<option value="Chronic">Chronic</option>

</select>
</td>


<td>Associated_Disease<input type="text" name="DisDu"></td>
</tr>

<td>Ethinicity<select name="DisType">
<option value="Select">Indian</option>
<option value="Acute">European</option>
<option value="Chronic">oman</option>
<option value="Acquired">German</option>


</select>
</td>

<td>Source<textarea name="comments" cols="30" rows="4"></textarea><br></td>
</tr>
</table>

</div>
</br>
<div><center>
<input type="submit" name="submit">
</center></div></div>
</form>
<?php
if(isset($_POST['submit'])){
$conn = mysqli_connect('localhost','root','');
if (!$conn) {
die("Connection failed: " . mysqli_error());
}
echo "DB Connected successfully";
mysqli_select_db("tool",$conn);
$sql="INSERT INTO disease (Disease_id, Disease,SubDisease, Associated_Disease, Ethinicity,Source)
VALUES ('$_POST[Disease_ID]', '$_POST[Disease]','$_POST[SubDisease]', '$_POST[Associated_Disease]','$_POST[Ethinicity]', '$_POST[Source]')";
mysqli_query($sql,$conn);
mysqli_close($conn);
}
?>
</body>
</html>

最佳答案

如果您可以接受 SQL 注入(inject)和被黑客攻击,请继续并遵循此建议。如果没有,你最好继续阅读 prepared statements相反..

1) 除非你的配置文件列出了数据库,否则你需要在你的构造函数中指定它:

$conn = mysqli_connect('localhost','root','', 'myDatabaseName');

2) 当在字符串中使用数组索引时,你应该用 curl 将它们包裹起来:

$sql="INSERT INTO  disease(Disease_id,Disease,SubDisease,Associated_Disease,Ethinicity,Source) VALUES           ('{$_POST[Disease_ID]}','{$_POST[Disease]}','{$_POST[SubDisease]]','{$_POST[Associated_Disease]}','{$_POST[Ethinicity]}','{$_POST[Source]}')";

关于php - 在php和mysql中将数据插入数据库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37117087/

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