我遇到了问题，无法找到如何在表列结果中创建超链接，然后单击打开另一个填充了所有字段（文本框）的页面。想象一下，当点击一个ID时，我从column_ID=ID的表中选择*。有办法做到这一点吗？谢谢
向致以最良好的问候
我不完全确定你在问什么，但这可能会对你有所帮助。
首先制作一个Javascript。
<script type="text/JavaScript">
function selectID() {
var ID = document.getElementById("ID").value;
document.location.href ="yoursite.php?ID="+ID;
}
</script>
然后连接到数据库，例如通过更改变量$value来查询表中的链接ID（或更多）。
<?php
//Connect to database
mysql_connect("host", "user", "pass");
mysql_select_db("db_name");
$value = 'something';
$ID = $_GET['ID'];
if (!$ID) {
$ID = 0;
}
if ($ID == 0) {
$query = "SELECT * FROM table WHERE `column_1` = '$value'";
$result = mysql_query($query);
echo "<table>";
while($myrow = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>";
echo "<a href='"yoursite.php?ID=" . $myrow[column_id] . "'" onclick='"selectID();'">ID</a>";
echo "</td>";
echo "</tr>";
}
echo "</table>";
}
elseif ($ID > 0) {
$query2 = "SELECT * FROM table WHERE `column_id` = '$ID'";
$result2 = mysql_query($query2);
while($myrow2 = mysql_fetch_array($result2)) {
$value1 = $myrow2['column_1'];
$value2 = $myrow2['column_2'];
}
echo "<form type='"GET'" action='"$PHP_SELF'">";
echo "<input type='"text'" id='"ID'" name='"ID'" value='"$ID'"><br>";
echo "<input type='"text'" id='"value1'" name='"value1'" value='"$value1'"><br>";
echo "<input type='"text'" id='"value2'" name='"value2'" value='"$value2'"><br>";
echo "<input type='"hidden'" id='"search'" name='"search'" value='"searching'">";
echo "<input type='"submit'" id='"submitbutton'" name='"submitbutton'" value='" Search '">";
echo "</form>";
}
?>