您好我已经对此进行了长时间的搜索,但没有找到解决办法。
我的代码如下:
$link = mysqli_connect("localhost",".........","...........",".........") or die("Error " . mysqli_error($link));
$ctime = time();
$check = "SELECT * FROM thread WHERE forumid='48' AND visible='1' ORDER BY lastpost DESC LIMIT 1" or die("Error in the consult.." . mysqli_error($link));
//execute the query.
$rc = mysqli_query($link, $check);
while($rows = $rc->fetch_assoc()){
$pid = $rows['firstpostid'];
$query = "SELECT * FROM dropouts WHERE date <= $ctime" or die("Error in the consult.." . mysqli_error($link));
//execute the query.
$result = mysqli_query($link, $query);
$row_cnt = $result->num_rows;
while($row = $result->fetch_array())
{
$date = $row['date'];
$user = $row['username'];
$sql = "DELETE FROM dropouts WHERE date = $date" or die("Error in the consult.." . mysqli_error($link));
//execute the query.
$done = mysqli_query($link, $sql);
//////////////////////////////////////////////////////////////////////////\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
$check1 = "SELECT * FROM post WHERE postid='$pid'" or die("Error in the consult.." . mysqli_error($link));
//execute the query.
$rc1 = mysqli_query($link, $check1);
while($row1 = $rc1->fetch_array())
{
$text = $row1['pagetext'];
echo str_replace($user, "", $text);
}
}
}
问题在于,当我像那样运行时,我没有出局。
如果我运行它以便while循环不在彼此中,我会得到输出,但脚本只对一行/一行进行。
有谁知道如何解决这个问题?
我读到它应该可以工作,但这只是工作......
由于
答案 0 :(得分:0)
你试过回应它吗?可能是查询返回0或为空!
作为php的初学者,你可能会遇到这些问题。了解问题的最佳方法是ECHO&amp;出口。通过执行此操作检查每个步骤,您可以更好地了解问题。
在上面的问题中,不是编写整个代码,而是首先尝试检查代码的一小部分并记住:
最好的编译器/解释器介于两只耳朵之间!!