[英]select query and while loop doesn't return all rows correctly
你可以解釋一下為什么當我嘗試這個查詢時它只返回給我一半的行?
例如,如果$ record由4個值組成,則只從DB獲取第1行和第3行。
怎么了?
$query=mysql_query("SELECT * FROM ".DB_PREF."books WHERE book_id IN ('".$records."')");
while($fetch=mysql_fetch_assoc($query))
{
global $book_id, $book_title, $book_description, $book_author_id, $book_author_name, $book_author_surname;
$book_id=$fetch['book_id'];
$book_title=$fetch['book_title'];
$book_description=$fetch['book_description'];
$book_author_id=$fetch['book_author_id'];
$query=mysql_query("SELECT * FROM ".DB_PREF."profiles WHERE user_id='".$book_author_id."'");
$fetch=mysql_fetch_assoc($query);
$book_author_name=$fetch['user_name'];
$book_author_surname=$fetch['user_surname'];
getModule('htmlmodule...');
}
你是否在循環中覆蓋你的$ fetch變量? 也許試試:
$fetch2=mysql_fetch_assoc($query);
或者,更好的是,在SQL中使用連接:
$query=mysql_query("SELECT * FROM ".DB_PREF."books LEFT JOIN ".DB_PREF."profiles ON book_author_id = user_id WHERE book_id IN ('".$records."')");
然后從單個查詢中獲取您想要的所有內容。
我做什么
mysql_select_db($database_lolx, $lolx);
$query_Recordset1 = "SELECT * FROM person";
$Recordset1 = mysql_query($query_Recordset1, $lolx) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
<table border="1">
<tr>
<td>id</td>
<td>emal</td>
</tr>
<?php do { ?>
<tr>
<td><?php echo $row_Recordset1['id']; ?></td>
<td><?php echo $row_Recordset1['emal']; ?></td>
</tr>
<?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>
</table>
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