我正在从mysql表(jobs)中获取行。在这个提取过程中,我还从另一个表(accounts)中提取[以接收account api密钥,这一切都取决于ID_ASSOC附加到作业中的内容]:下面是代码
$sql = "SELECT * FROM jobs";
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
$sql = "SELECT * FROM accounts WHERE id_assoc='$job_poster_id'";
$query = mysqli_query($db_conx, $sql);
while($rows = mysqli_fetch_assoc($query)){
$username = $rows['twitter_username'];
$consumer_key = $rows['consumer_key'];
$consumer_secret = $rows['consumer_secret'];
$access_token = $rows['access_token'];
$access_token_secret = $rows['access_token_secret'];
}
echo $job_poster_id ;
echo "<br/>";
echo $twitter_username;
echo "<br/>";
echo "----------------------------------";
echo "<br/>";
}
输出:
specific-message
4
admin
----------------------------------
当我这样做时,我只得到一行输出。。我似乎不知道为什么。我希望上面的输出重复多次,因为它有行,而且它只做一行(代码中有账户提取)。然而,当我在没有内部提取(accounts fetch)的情况下执行此操作时,它会根据需要返回多行。为什么会这样?(下面是没有账户提取的示例代码):
$sql = "SELECT * FROM jobs";
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
echo $job_poster_id ;
echo "<br/>";
echo "----------------------------------";
echo "<br/>";
}
输出:
specific-message
4
----------------------------------
specific-message
1
----------------------------------
specific-message
2
----------------------------------
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
$sql = "SELECT * FROM accounts WHERE id_assoc='$job_poster_id'";
$query = mysqli_query($db_conx, $sql);
问题是,您正在将$query
用于内部和外部查询。
当内部查询运行并逐步通过循环时,它将迭代到结果集的末尾;当外部while循环运行时,mysqli_fetch_assoc($query)
返回false,因为您已经处于结果集的末尾,只是不是您期望的结果集。
可以通过重命名其中一个$query
变量来解决此问题。