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 06:09:16
-- notice the error in first query
SELECT MIN(Bale_ID), from data WHERE (fchmitimestamp LIKE CONCAT(@year,'%',@month,'%',@day,'_________'))
SELECT MIN(Incoming_Moisture) from data WHERE (fchmitimestamp LIKE CONCAT(@year,'%',@month,'%',@day,'_________'))使用var_dump($selection)而不是echo $selection来查看自己
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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