我希望这些数据显示所有结果,在查询中我得到 129 个结果。但是当我在页面上显示它时,我只得到一行。我以前使用过非常相似的代码来获得多个结果,所以我知道这很简单,但我就是不明白。任何想法将不胜感激!
<?php
$sql = "SELECT SUM(datamb) AS value_sum FROM maindata GROUP BY phonenumber";
$sql1 = "select dataplan as currentplan from maindata GROUP BY phonenumber";
$sql2 = "SELECT DISTINCT phonenumber AS value_sum1 FROM maindata";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
$result1 = mysql_query($sql2);
if (!$result1) {
echo "Could not successfully run query ($sql1) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result1) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
$result2 = mysql_query($sql2);
if (!$result2) {
echo "Could not successfully run query ($sql2) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result2) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($result)){
echo "<TABLE id='display'>";
echo "<td><b>Data Usage This Period: ". ROUND ($row["value_sum"],2) . "MB</b></td> ";
}
while ($row1 = mysql_fetch_assoc($result1)){
echo "<TABLE id='display'>";
echo "<td><b>Data Plan: ". $row1["currentplan"] . "</b></td> ";
}
while ($row2 = mysql_fetch_assoc($result2)){
echo "<TABLE id='display'>";
echo "<td><b>Phone Number: ". $row2["value_sum1"] . "</b></td> ";
}
?>
根据建议更新 - 非常有帮助,谢谢,我非常接近,所有值都是正确的,但我无法将它们放在同一张表中,想法?