Question

我在一个数据库中有表，一个是外键到另一个。我的问题是我试图根据用户名调用存储在一个表中的信息，该用户名链接存储在另一个表中的2个表。这是我的PHP，请注意，我在数据库和php上非常新鲜，所以减少了一些空闲。这是我的代码：

<?php

$loaduser= $_SESSION['username'];
$loaduser_conn= @mysql_connect("DB_NAME","DB_USER","DB_PASS");

mysql_select_db("user_register") or die ("Couldn't find user database.");

$gal_result= mysql_query("SELECT shoot_name FROM images WHERE username='$loaduser'") or die (mysql_error());

while($row = mysql_fetch_assoc($gal_result,MYSQL_ASSOC))
{
    foreach($results['shoot_name'] as $result)
    {
        echo $result['shoot_name'], '<br>';

        if(mysql_num_rows($gal_result) !=1) {
            die("No galleries found for this user.");
        }

    }
}

Answer 1

$ results必须在tryin获取其内容之前获取。
在尝试获取其内容时，您获取$ row而不是$ results
您正在使用mysql，不推荐使用mysql，请在将来尝试使用mysqli或PDO。

无论如何都要试试。

<?php
$loaduser= $_SESSION['username'];

$loaduser_conn= @mysql_connect("DB_NAME","DB_USER","DB_PASS");
mysql_select_db("user_register") or die ("Couldn't find user database.");

$gal_result= mysql_query("SELECT shoot_name FROM images WHERE username='$loaduser'") or die (mysql_error());

$results = mysql_fetch_assoc($gal_result,MYSQL_ASSOC)

while($results = mysql_fetch_assoc($gal_result,MYSQL_ASSOC)){
  foreach($results['shoot_name'] as $result) {

  echo $result['shoot_name'], '<br>';

  if(mysql_num_rows($gal_result) !=1){
    die("No galleries found for this user.");
  }

  }
}
?>

Answer 2

尝试重写这个未经测试的代码段（如果您想继续使用已弃用的代码;）

$loaduser= $_SESSION['username'];
$loaduser_conn= @mysql_connect("DB_NAME","DB_USER","DB_PASS");
mysql_select_db("user_register") or die ("Couldn't find user database.");
$gal_result= mysql_query("SELECT shoot_name FROM images WHERE username='$loaduser'") or die (mysql_error());
if(mysql_num_rows($gal_result)==0){ // returns number of rows so if 0 there are no rows
    die("No galleries found for this user.");
}
while ($row = mysql_fetch_array($gal_result)) { // "while" is already your loop. No need for the foreach within.
    echo $row['shoot_name'], '<br>';
}

如果您想从多个表中选择一些参考，请尝试以下查询：

$gal_result= mysql_query("SELECT shoot_name FROM images AS a LEFT JOIN your_other_table AS b ON a.username = b.username WHERE a.username='$loaduser'") or die (mysql_error());

像任何人一样，我会建议您使用更新的代码。有关详细信息，请参阅此处：http://php.net/manual/de/function.mysql-query.php

Answer 3

您可以谷歌“PHP PDO教程”并找到许多资源。这是一篇非常清晰且写得很好的文章：例如：https://phpdelusions.net/pdo

这是一个更好的例子：

<?php

$loaduser = $_SESSION['username'];

// use PDO, not the deprecated mysql extension
$dsn = "mysql:host=localhost;dbname=user_register";
$conn = new PDO($dsn, "DB_USER", "DB_PASS");

// set exception errmode, so code dies automatically if there's an error
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

// use a parameterized query instead of concatenating variables into SQL 
$sql = "SELECT shoot_name FROM images WHERE username = ?";
$stmt = $conn->prepare($sql);
$stmt->execute([$loaduser]);

// fetch all into an array of results
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);

// check the count of the results with < 1 instead of != 1
// in case it's 2 or more
if ($stmt->rowCount() < 1) {
    die("No galleries found for this user.");
}

// loop through results
foreach($results as $row) {
    // use dot for string concatenation instead of comma. 
    echo $row["shoot_name"] . "<br>";
}

回声结果没有显示

3 个答案: