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SQL语句转换为PHP变量时的情况_Php_Sql_Variables - Fatal编程技术网

SQL语句转换为PHP变量时的情况

SQL语句转换为PHP变量时的情况,php,sql,variables,Php,Sql,Variables,我有一个问题: 我有一个SQL语句: SELECT Amount FROM cicmpy INNER JOIN (SELECT DealerCode, sum(Amount) as Amount FROM (SELECT DealerCode, LOG as LogtLimit,TemporaryCreditLimit as TemporaryCreditLimit, CASE WHEN TemporaryCreditLimit>0 then Te

我有一个问题:

我有一个SQL语句:

SELECT Amount FROM cicmpy 
  INNER JOIN (SELECT DealerCode, sum(Amount) as Amount FROM 
    (SELECT DealerCode, LOG as LogtLimit,TemporaryCreditLimit as TemporaryCreditLimit,
            CASE WHEN  TemporaryCreditLimit>0 then TemporaryCreditLimit ELSE LOG END as Amount 
            FROM LOG_DATA WHERE Status = 1 group by DealerCode) x
    on x.DealerCode=debcode where debnr is not NULL and ltrim(debcode) = '21021287'`
我想把这个语句转换成php$query变量

像这样试试,它会解决你的错误


$query=从cicmpy内部联接选择DealerCode中选择金额,从选择DealerCode中选择sumAmount作为金额,记录为LogtLimit,临时CreditLimit作为临时CreditLimit,当TemporaryCreditLimit>0时,则TemporaryCreditLimit ELSE日志结束为日志数据中的金额,其中状态=1,按x上的DealerCode x分组。DealerCode=debcode,其中debnr不为NULL,ltrimdebcode='21021287'。。。不或者您是在问如何将此语句的结果放入变量中?否,因为当我调用sqlsrv_query$conn$query时;在调用sqlsrv_fetch_数组之后,出现了如下错误:警告:sqlsrv_fetch_数组希望参数1是资源,布尔值在第13行C:\xampp\htdocs\log\u data\test.php中给出:facepalm:当错误显示为参数1时,您真的认为这是第二个不好的参数吗?可能您的连接不好,booleand是假的。请创建一个var_dump$conn probabli为FALSE,可能您只是没有连接。我有这样一个查询:从cicmpy内部连接选择DealerCode选择金额,从选择DealerCode选择sumAmount作为金额,记录为LogtLimit,临时CreditLimit作为临时CreditLimit,案例当TemporaryCreditLimit>0时,TemporaryCreditLimit ELSE日志结束为日志数据中的金额,其中Status=1根据DealerCode x对x进行分组。DealerCode=debcode,其中debnr不为NULL,ltrimdebcode='21021287'将查询放在他编写的地方并将查询放在那里。如果出现错误,可能是因为您没有创建到数据库的连接。首先测试一个非常简单的查询,以确保您可以进行查询,然后为它提供您的复杂查询。
$sql = "Put your query";    //put your query in php variable
$result = mysql_query($sql) or die(mysql_error());
while($row=mysql_fetch_array($result)) {
do stuff...
}