[英]mysql select, count, left join and show
我必須進行mysql詢問,並且必須按日期計數一些具有相同值的單元格,並從其他表中加入其他信息。
表格1
col1 | col2 | col3
ALEX | today | finished
JOHN | today | finished
TIM | today | finished
JOHN | today | unfinished
JOHN | today | finished
TIM | tommorow | finished
表2
col4 | col5
ALEX | mail1@website.tld
JOHN | mail1@website.tld
TIM | mail1@website.tld
我嘗試了這段代碼:
sql="
SELECT col2
, col1
, col3
, COUNT(*)
FROM table1
LEFT
JOIN table2
ON table1.task_cine = table2.col4
WHERE table1.col3='finished'
GROUP
BY col1";
$result = mysql_query($sql) or die($sql.mysql_error());
while($rows=mysql_fetch_array($result))
{
echo "user : ";
echo $rows['col1'];
echo ", on date: ";
echo $rows['col2'];
echo " have ";
echo $row['COUNT(*)'];
echo " finished, and have email address";
echo $rows['col5'];
}
你能幫我嗎? 我試圖了解php,但是很難。
嘗試:-
mysql_fetch_assoc($ result)
代替mysql_fetch_array($ result)
要么
將索引與數組參數一起用於mysql_fetch_array($ result)
如$ rows [0],$ rows [1]
嘗試這個:
$sql="
SELECT col1
, col2
, col5
, COUNT(*)
FROM table1
LEFT
JOIN table2
ON table1.task_cine = table2.col4
WHERE table1.col3='finished'
GROUP
BY col1";
$result = mysql_query($sql) or die($sql.mysql_error());
while($rows=mysql_fetch_array($result))
{
echo "user : ";
echo $rows['col1'];
echo ", on date: ";
echo $rows['col2'];
echo " have ";
echo $rows['COUNT(*)'];
echo " finished, and have email address";
echo $rows['col5'];
}
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