Question

这个问题已在本网站上被问过几次，但是这些解决方案都没有对我有用。

我在查询中运行查询：

 //this query is to count variables
$sql = "SELECT variabletype, COUNT(variabletype) AS value_occurrence FROM variable GROUP BY variabletype ORDER BY value_occurrence DESC";
$vars_query = mysqli_query($con,$sql) or die(mysqli_error($con));
$pos = array();

while($vars = mysqli_fetch_array($vars_query, MYSQLI_ASSOC))
    {

  //each variable has many different values, some positive, some negative, in this query I am trying to understand of the values found, how many of them are positive and pushing the result into an array. 
 $times_positive_qry = mysqli_query($con, "SELECT variable.variabletype, COUNT(value.valueid) AS positive_occurrence FROM variable INNER JOIN value On variable.variableid=value.variableid WHERE variable.variabletype = '" .$vars['variabletype']. "' AND value.valuelift>0.00 AND value.valuesignificant=1 GROUP BY variable.variabletype ORDER BY positive_occurrence DESC");
 $times_positive = mysqli_fetch_array($times_positive_qry, MYSQLI_ASSOC);
 $pos[] = $times_positive['positive_occurrence'];

}

我的结果是：

var s1 = ["10","3","2","3","4","5","3","3","2","1",null,"2","2",null,"4","3","3","2","2","2","1","1","1",null,"1",null,null,null,"1","1",null,"1","1",null,"3",null,"1","1","1",null]

您会注意到某些结果以 null 的形式返回，这意味着特定变量没有正值。无论如何，我可以用0＆替换空值？我已尝试IFNULL和COALESCE，但他们没有工作。

最终我需要空值为0，因为我正在绘制数组图形，如果我有空值，图形将不会显示。

有什么想法吗？

Answer 1

使用IFNULL功能，如

IFNULL(your_column,0)

根据您的评论，您错误地执行了coalesce。它应该像下面一样。原因如果valueid为null，那么COUNT(value.valueid)无论如何都会为0（尝试将样本视为select count(null) as total）。

SELECT IFNULL(variable.variabletype,0), <-- Change Here
COUNT(value.valueid) AS positive_occurrence
FROM variable 
INNER JOIN value 
On 
variable.variableid=value.variableid 
WHERE variable.variabletype = '" .$vars['variabletype']. "' 
AND 
value.valuelift>0.00 
AND 
value.valuesignificant=1
GROUP BY variable.variabletype 
ORDER BY positive_occurrence DESC

Answer 2

而不是

"SELECT variable.variabletype, COUNT(value.valueid)..."

"SELECT variable.variabletype, isnull(COUNT(value.valueid),0)"

Answer 3

尝试使用isnull函数

ISNULL（YOUR_VAR，0）

http://technet.microsoft.com/en-us/library/ms184325.aspx

Answer 4

你可以使用ISNULL（）函数

ISNULL(COUNT(value.valueid),0)

Answer 5

如果你想要你可以手动浏览你的php数组并进行字符串替换，或者你可以访问数据库，你可以做类似的事情 "update variable set value=0 where value is NULL;"

Answer 6

真正的问题在于第二个问题：

$vars_query = mysqli_query($con,$sql) or die(mysqli_error($con));
$pos = array();

while($vars = mysqli_fetch_array($vars_query, MYSQLI_ASSOC)) {
    //each variable has many different values, some positive, some negative, in this query I am trying to understand of the values found, how many of them are positive and pushing the result into an array. 
    $times_positive_qry = mysqli_query($con, "SELECT variable.variabletype, COUNT(value.valueid) AS positive_occurrence FROM variable INNER JOIN value On variable.variableid=value.variableid WHERE variable.variabletype = '" .$vars['variabletype']. "' AND value.valuelift>0.00 AND value.valuesignificant=1 GROUP BY variable.variabletype ORDER BY positive_occurrence DESC");         $times_positive = mysqli_fetch_array($times_positive_qry, MYSQLI_ASSOC);
    $pos[] = $times_positive['positive_occurrence'];
}

问题是您的WHERE子句可以消除所有结果。这意味着整个结果集可以为null。 您将获得零行的结果集。您需要使用LEFT JOIN来确保始终获得一行。

$times_positive_qry的查询字符串应如下所示：

"SELECT variable.variabletype, 
    COUNT(value.valueid) AS positive_occurrence 
FROM variable 
LEFT JOIN value 
    ON variable.variableid = value.variableid 
WHERE variable.variabletype = '" .$vars['variabletype']. "' 
    AND ((value.valuelift > 0.00 AND value.valuesignificant = 1)
        OR
    (value.valuelift IS NULL AND value.valuesignificant IS NULL))    
GROUP BY variable.variabletype 
ORDER BY positive_occurrence DESC"

注意：如果存在变量可以value.valuelift IS NULL AND value.valuesignificant IS NULL为真的合法情况，则会遇到问题。您可能需要添加value.variableid IS NULL ...在不查看数据的情况下很难说。

将Null从SQL更改为0

6 个答案: