Question

我必须进行一次mysql审讯，我必须计算一些具有相同值的单元格，按日期计算，并从左表中加入其他信息。

表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

我试过这段代码：

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'];
}

你能帮帮我吗？我试图理解php，但很难。

Answer 1

尝试： -
mysql_fetch_assoc（$结果）

而不是 mysql_fetch_array（$结果）

或

使用带有mysql_fetch_array（$ result）

的数组参数的索引

为$ rows [0]，$ rows [1]

Answer 2

试试这个：

$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'];
}

mysql select，count，left join和show

2 个答案: