我是php和mysql的新手，我想了解如何将数据从数据库字段传输到php表单。我在一个表上显示了数据，在最后一列上，我有一个按钮（赞助商），您可以在其中选择该行，一个特定的字段将显示在一个表单上，一旦您选择按钮（赞助者），该表单就会打开。我有三个显示数据的脚本（needed.php），其中按钮发送所选的id/No字段（add.php），下面的表单是要发送数据的表单（details.php）。请帮助了解如何进行。
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>
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，要验证POST/GET变量，并保护自己免受add.php页面上SQL注入的影响，还有很多工作要做——我会将add.php和details.php脚本结合起来，但这是基于您的示例的简单化视图，将它们分开可能还有其他原因。
php-mysql函数也被弃用(http://www.php.net/mysql_query)现在考虑将脚本更新为mysqli_query或PDO，并阅读SQL注入预防：如何在PHP中防止SQL注入？