[英]For each / query doesn't give the right value
我的数据库中有一个表。 我想从leave
表中获取ltotal
值,然后计算所有ltotal
。 这是我使用的查询和代码:
$annual_query = pg_query("select ltotal from leave where lapplicant='adam' and ltype=2");
$annual_result = pg_fetch_array($annual_query);
if (pg_num_rows($annual_query) > 0) {
foreach ($annual_result as $data) {
$total_annual = $total_annual + $data;
}
print($total_annual);
}
表离开中有3条记录,其中lapplicant='adam'
和ltype=2
。
每个ltotal
为1。
当我尝试运行print($total_annual)
,结果为2(必须为3)。
然后我尝试print_r($annual_result['ltotal']
),结果仅为1(必须为1,1,1)。
谁能帮我? 谢谢。
泰米尔人指出了眼前的问题。 但是-为什么不只在SQL中这样做呢?
select sum(ltotal)
from leave
where lapplicant='adam' and ltype=2
pg_fetch_array()
仅返回带有数字键和关联键的行(遍历时,相同值两次)。 您应该使用pg_fetch_all()
并遍历或在连续的行上使用while循环。
$total_annual = 0;
$annual_query = pg_query("select ltotal from leave where lapplicant='adam' and ltype=2");
while ($row = pg_fetch_array($annual_query)) {
$total_annual = $total_annual + $row['ltotal'];
}
print($total_annual);
要么
$total_annual = 0;
$annual_query = pg_query("select ltotal from leave where lapplicant='adam' and ltype=2");
$annual_result = pg_fetch_all($annual_query);
foreach ($annual_result as $row) {
$total_annual = $total_annual + $row['ltotal'];
}
print($total_annual);
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