目前,$selection输出以下内容: MIN(Bale_ID),MIN(Incoming_Moisture),这正是它应该输出的内容(它们是来自另一个表的名称)。但是,当我将$selection放入mysql_query $data1中时,它似乎只是读取最后一个值(MIN(Incoming_Moisture)),并且只显示结果。如何让查询读取$selection中的整个元素数组?谢谢你!!
while ($row1 = mysql_fetch_array($fieldnames1)) {   
$fields = $row1['fields1']; 
$explode = explode(',',$fields);
if ($row1) {
    for ($i=0; $i<$minrows; $i++) {
        if ($i<$minrows-1){
            $comma = ", ";
        }
        else {
            $comma = "";
        }
        //$selection = "MIN(".$explode[$i].")".$comma;
        //echo $selection;
        $data1 = mysql_query("SELECT MIN(".$explode[$i].")".$comma." from data WHERE (fchmitimestamp LIKE CONCAT(@year,'%',@month,'%',@day,'_________'))");
        $all1 = mysql_num_fields($data1); //return # of columns; for some reason is returning "1" right now.
        while ($row2 = mysql_fetch_array($data1)) {
            for ($col=0; $col<$all1; $col++) {
                echo $all1;
                echo "<td>Min: " . $row2[$col] . "</td>";
            }
            echo "</tr>";
        }
    }       
}
}
echo "</table>";发布于 2011-07-22 05:20:08
看看代码中的操作顺序:
loop {
     ... fetch data ...
     ... assign results to $data1 ...
}在您的循环中,您不会输出或保存在$data1中获得的结果,因此循环的每次迭代都会覆盖前一次迭代的结果-换句话说,只有最后一次迭代的结果将被存储。
https://stackoverflow.com/questions/6782913
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