Question

这是另一个指向update.php文件的php文件的链接。我从这个链接获得了链接的ID。现在我想使用该id从数据库中检索数据我该怎么做？

echo '<a href="update.php?id= "'.$row['id'].'">Modify</a>';

现在这是我的update.php文件

echo "<form method=\"POST\" action=\"\">\n"; 
include_once 'dbconnect.php';
$id= $_GET['id'];
$query  = "SELECT * FROM promoter where id=$id ";// this code is not retrieving value from database
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{   $user_id=$row['user_id'];
$full_name=$row['full_name'];
$qualification=$row['qualification'];
$locality=$row['locality'];
$description=$row['description'];
}

echo "<td>User Id</td><td> <input type=\"text\" name=\"user_id\" value=\"$user_id\"></td>\n"; 
echo "</tr>\n"; 
echo "<tr>\n"; 
echo "<td>Full Name</td><td> <input type=\"text\" name=\"full_name\" value=\"    $full_name\"></td>\n"; 
echo "</tr>\n"; 
echo "<tr>\n"; 
echo "<td>Qualification</td><td> <input type=\"text\" name=\"qualification\" value=\" $qualification\"></td>\n"; 
echo "</tr>\n"; 
echo "<tr>\n"; 
echo "<td>Locality</td><td> <input type=\"text\" name=\"locality\" value=\" $locality\"></td>\n"; 
echo "</tr>\n"; 
echo "<tr>\n"; 
echo "<td>Description</td><td> <input type=\"text\" name=\"description\"     value=\" $description\"></td>\n"; 
echo "</tr>\n";

echo "</form>\n";

现在我想从数据库中检索数据，其中id =我从链接获得的id值，以便我可以根据id显示值

Answer 1

首先你检查id检索到update.php检查。

echo $id;

如果显示，则必须检查是否使用SQL查询从数据库中检索任何值。

print_r($result);

用下面的代码替换

echo "<form>\n"; 
include_once 'dbconnect.php';
$id= $_GET['id'];
$query  = "SELECT * FROM promoter WHERE id='$id'";
$result = mysql_query($query);
if ($row = mysql_fetch_array($result))
{
    $user_id=$row['user_id'];
    $full_name=$row['full_name'];
    $qualification=$row['qualification'];
    $locality=$row['locality'];
    $description=$row['description'];

    echo "<td>User Id</td><td> <input type=\"text\" name=\"user_id\" value=\"$user_id\"></td>\n"; 
    echo "</tr>\n"; 
    echo "<tr>\n"; 
    echo "<td>Full Name</td><td> <input type=\"text\" name=\"full_name\" value=\"    $full_name\"></td>\n"; 
    echo "</tr>\n"; 
    echo "<tr>\n"; 
    echo "<td>Qualification</td><td> <input type=\"text\" name=\"qualification\" value=\" $qualification\"></td>\n"; 
    echo "</tr>\n"; 
    echo "<tr>\n"; 
    echo "<td>Locality</td><td> <input type=\"text\" name=\"locality\" value=\" $locality\"></td>\n"; 
    echo "</tr>\n"; 
    echo "<tr>\n"; 
    echo "<td>Description</td><td> <input type=\"text\" name=\"description\"     value=\" $description\"></td>\n"; 
    echo "</tr>\n";
    echo "</form>\n"; 
}

Answer 2

这是您的解决方案工作代码是。

  echo '<a href="update.php?id='.$row['id'].'">Modify</a>';

用此行替换上面的代码。实际上你的错误是你在URL完成之前已经完成了“，所以你的动态传递的值不被视为参数我只是删除那个双（"）。希望它能帮到你。

从id值的数据库中检索数据

2 个答案: