我是 php 和 mysql 的新手，我想要一些有关如何将数据从数据库字段传输到 php 表单的帮助。我已将数据显示在表格上，并且在最后一列上我有一个按钮(赞助商)，其中您可以选择该行，特定字段将显示在选择按钮(赞助商)后打开的表单上。我有三个显示数据的脚本(needy.php)，其中按钮发送 id/No 字段选择(add.php)，然后将要发送数据的表单(details.php)下面是代码。请帮助了解如何进行操作。
needy.php
<?php
$username = "egesachi_baby";
$password = "babyclass" ;
$database = "egesachi_babyclass";
$server = "localhost";
$db_handle = mysql_connect($server,$username,$password);
$db_found = mysql_select_db($database)or die ("cannot connect");
if ($db_found) {
$sql = "SELECT * FROM needy";
$result= mysql_query($sql);
echo "<table border='1'>
<tr>
<th>Photo</th>
<th>Name</th>
<th>Age</th>
</tr>";
while($db_field = mysql_fetch_assoc($result)) {
echo "<tr>";
echo "<td>" . '<img src="data:image/jpeg;base64,' .
base64_encode( $db_field['Photo']) .
'" width="280" height="280" />'; "</td>";
echo "<td>" . $db_field['Name']."</td>";
echo "<td>" . $db_field['Age']."</td>";
echo "<td>";
echo "<form action='add.php' method='post'>";
echo "<input type='hidden' name='pupilno' value='".$db_field['No']."'>";
echo "<input type='submit' value='sponsor' />";
echo "</form>";
echo "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($db_handle);
}
else {
print "database not found";
mysql_close($db_handle);
}
?>
add.php
<?php
session_start();
$username = "egesachi_baby";
$password = "babyclass" ;
$database = "egesachi_babyclass";
$server = "localhost";
$db_handle = mysql_connect($server,$username,$password);
$db_found = mysql_select_db($database)or die ("cannot connect");
if ($db_found) {
if(isset($_POST['sponsor'])) {
$id = $_POST['pupilno'];
$sql = "SELECT Name FROM needy WHERE No= '$_POST['pupilno']'";
$result= mysql_query($sql);
while ( $db_field = mysql_fetch_assoc($result) ) {
$sponsored = $db_field['Name'] . "<BR>";
}
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
}
?>
details.php
<html>
<body>
<form action="paypal.php" method="post">
Pupil No: <input type="text" name="pupilid"><p/>
pupil Name: <input type="text" name="sponsored"><p/>
Your Name: <input type="text" name="name"><p/>
Your Email :<input type="text" name="email"><p/>
Phone No: <input type="text" name="phone"><p/r>
Country: <input type="text" name="country"><p/>
Contribution:<input type="text" name="con"><p/>
<input type="submit" value="Send it!">
</form>
</body>
</html>
最佳答案
试试这个:
add.php
...
while ( $db_field = mysql_fetch_assoc($result) ) {
$sponsored = $db_field['Name'];
}
header('Location: add.php?pupil='.$_POST['pupilno'].'&name='.$sponsored );
...
然后
details.php
...
<form action="paypal.php" method="post">
Pupil No: <input type="text" name="pupilid" value="<?php $_GET['pupil'] ?>"><p/>
pupil Name: <input type="text" name="sponsored" value="<?php $_GET['name'] ?>"><p/>
Your Name: <input type="text" name="name"><p/>
Your Email :<input type="text" name="email"><p/>
Phone No: <input type="text" name="phone"><p/r>
Country: <input type="text" name="country"><p/>
Contribution:<input type="text" name="con"><p/>
<input type="submit" value="Send it!">
</form>
...
然而，还需要做更多的事情来验证 POST/GET 变量并保护自己免受 add.php 页面上的 SQL 注入(inject) - 我会结合 add.php 和details.php 脚本，但这是基于根据您的示例的简单化观点，可能还有其他原因将它们分开。
php mysql 函数也已被弃用 ( http://www.php.net/mysql_query )，现在考虑将脚本更新为 mysqli_query 或 PDO 并阅读 SQL 注入(inject)预防: How can I prevent SQL injection in PHP?
关于php - 将数据库项目添加到 php 表单，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20440340/