Question

我试图通过单击表中的字段来调用phpmyadmin中存储的所有表信息。但我遇到的ID列是主键的问题

数据控制文件：

注意：未定义的索引：第52行的C：\ xampp \ htdocs \ project \ project01 \ datacontrol.php中的ID

详细文件：

警告：mysqli_fetch_assoc（）期望参数1为mysqli_result，在第22行的C：\ xampp \ htdocs \ project \ project01 \ details.php中给出的布尔值

这是表格代码：

while($record = mysqli_fetch_array($myData)) {
    echo "<form action=datacontrol.php method=post>";
    echo "<tr>";
    echo "<td>" . "<a name style='color:black; text-decoration: none; background-color: none;' 
href='details.php?id=<?php echo $record[id]?>'>" . $record['Project_Name'] ."</a>" . " </td>";
    echo "<td>" . "<input type=text name=type value="
        . $record['Type'] . " </td>";
    echo "<td>" . "<input type=text name=cost value="
        . $record['Cost'] . " </td>";
    echo "<td>" . "<input type=hidden name=hidden value=" . $record['Project_Name'] . " </td>";
    echo "<td>" . "<input type=submit name=update value=Update" . " </td>";
    echo "<td>" . "<input type=submit name=delete value=Delete" . " </td>";

    echo "</tr>";
    echo "</form>";
}
echo "<form action=datacontrol.php method=post>";
echo "<tr>";
echo "<td><input type=text name=uprojectname></td>";
echo "<td><input type=text name=utype></td>";
echo "<td><input type=text name=ucost></td>";
echo "<td></td><td>" . "<input type=submit name=add value=  +Add  " . " </td>";
echo "</form>";
echo "</table>";
mysqli_close($con);

?>

这是我希望在单击后显示的详细信息页面：

<?php
$con = mysqli_connect("localhost", "root", "", "project");
if (!$con) {
    //display the reason for mysql connection error.
    die ("Connection Failed: " . mysqli_connect_errno());
}

mysqli_select_db($con, "project");

$id = $_GET['id'];

$query = 'SELECT * FROM projects WHERE id='.$id;
$result= mysqli_query($con, $query);
$record = mysqli_fetch_assoc($result);

?>



<h5><?php echo $record['projectname'];?></h5>
<br/>
<p><?php echo $record['type'];?></p>
<br/>
<?php echo $record['cost'];?>
<br/>
<?php echo $record['details'];?>

<?php
mysqli_close($con);

?>

Answer 1

您已经在PHP标记内回显了。因此，无需将PHP标记放入echo中。像这样改变这行

echo "<td>" . "<a name style='color:black; text-decoration: none; background-color: none;' 
    href='details.php?id='.$record['id'].'>'" . $record['Project_Name'] ."</a>" . " </td>";

获取表格详细信息的问题

1 个答案: