根据 join
优于嵌套查询的建议,我已将所有嵌套查询转换为 join
。但是,在转换为 join
后,我无法从 SQL 结果中将数据检索到我的数组中。
这是我的查询:
没有加入
$a="SELECT F_DATE, COUNT(F_DATE) as COUNT_F
from FWH
where FI_NAME IN
(
SELECT I_NAME from INS_W WHERE INSTANCE_ID IN
(
SELECT I_MAP_ID FROM T_MAP where T_MAP_ID =
(
SELECT T_ID FROM TWY WHERE T_NAME = 'abc'
)
)
)
AND F_DATE between '$S_D' AND '$E_D'
GROUP BY F_DATE";
加入
$a="SELECT t1.F_DATE AS DATE_F, COUNT(t1.F_DATE) as COUNT_F
from FWH t1
JOIN INS_W t2 ON(t1.FI_NAME = t2.I_NAME)
JOIN T_MAP t3 ON(t2.INSTANCE_ID = t3.I_MAP_ID)
JOIN TWY t4 ON(t3.T_MAP_ID = t4.T_ID)
WHERE t4.T_NAME = 'abc' AND
t1.F_DATE BETWEEN '$S_D' AND 'E_D'GROUP BY t1.F_DATE";
这是检索数据的 PHP 代码
$link = mysql_connect("ip", "user", "passs");
$dbcheck = mysql_select_db("db");
if ($dbcheck) {
$chart_array_1[] = "['F DATE','F COUNT']";
$result = mysql_query($a);
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_assoc($result)) {
$f_date=$row["DATE_F"];
$f_count=$row["COUNT_F"];
$chart_array_1[]="['".$f_date."',".$f_count."]";
}
}
}
mysqli_close($link);
直接在 MySQL 数据库上测试时,SQL 查询本身运行良好。
最佳答案
出于某种原因,当我使用连接时,我不得不使用行[0]、行[1] 等而不是使用列名获取值。我不明白这背后的原因。但是,这是我的唯一出路。以下代码适用于可能遇到与我类似情况的人。
$link = mysql_connect("ip", "user", "passs");
$dbcheck = mysql_select_db("db");
if ($dbcheck) {
$chart_array_1[] = "['F DATE','F COUNT']";
$result = mysql_query($a);
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_assoc($result)) {
$chart_array_1[]="['".$row[0]."',".$row[1]."]";
}
}
}
mysqli_close($link);
关于PHP/MYSQL - 数据检索,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30863643/