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Question

我从mysql中提取一些数据并计算匹配的出现次数（看似相当于WHERE foo='bar'）。但是，当我在PHP中循环数据时，我的计数远低于mysql中的计数。

MYSQL> SELECT COUNT(foo) FROM database.table WHERE foo='bar';

# PHP
while ($response = mysql_fetch_assoc($surveydata)){
    if ($response==='bar') {
        $bar++;
    }
}

数据可能包含一个或多个&，因此我只想匹配bar而不是bar & foobar。我怀疑mysql正在计算bar和bar & foobar，而php只计算bar而不是bar & foobar。 Php正在返回1210，并且mysql正在返回1783，因此手动计算以查看谁是正确的，这是非常实用的...

我google了一下，但是对于“mysql完全匹配”或“mysql完全相同”x，x

感到惊讶

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这是实际的sql

SELECT COUNT(`race`) FROM `database`.`table` WHERE `completion status`='complete';

和实际的php

mysql_query("SELECT `race`,`etcetera` FROM `database`.`table` WHERE `completion status`='complete';");

$demographics=array(
    "race"=>array(
        "White"=>array('consented'=>0,'partial'=>0,'completed'=>0),
        "Black"=>array('consented'=>0,'partial'=>0,'completed'=>0),
        "Hispanic"=>array('consented'=>0,'partial'=>0,'completed'=>0),
        "Asian"=>array('consented'=>0,'partial'=>0,'completed'=>0),
        "Pacific Islander"=>array('consented'=>0,'partial'=>0,'completed'=>0),
        "Multiracial"=>array('consented'=>0,'partial'=>0,'completed'=>0),
        "Other"=>array('consented'=>0,'partial'=>0,'completed'=>0)
    )
    //more
);

while ($dbrecord = mysql_fetch_assoc($surveydata)) {
    foreach ( $dbrecord as $dbfield=>$dbcellval ) {
        foreach ( $demographics as $demographic=>&$options ) {
            foreach ( $options as $option=>&$counter ) {
                if ( $option==="Multiracial" && strpos($dbcellval,'&') >0 && strpos($dbcellval,'&')!==false ) {
                    if ($dbrecord['consent']==="1"){
                        $demographics["race"]["Multiracial"]['consented']++;
                        if ($dbrecord['completion status']==="partial") {
                            $demographics["race"]["Multiracial"]['partial']++;
                        } // if
                        else if ($dbrecord['completion status']==="complete") {
                            $demographics["race"]["Multiracial"]['completed']++;
                        } // else if
                    } // if
                }
                else if ($option===$dbcellval){
                    if ($dbrecord['consent']==="1"){
                        $counter['consented']++;
                        if ($dbrecord['completion status']==="partial") {
                            $counter['partial']++;
                        } // if
                        else if ($dbrecord['completion status']==="complete") {
                            $counter['completed']++;
                        } // else if
                    } // if
                } // else if $option==$item
            } // foreach $options
        } // foreach $demographics
    } // foreach $dbrecord
} // while

来自SELECT race FROM database.table的数据如下：

White & Black
White
White & Asian
White & Asian & Black
Asian
Asian & Black
// etc

Answer 1

你可以这样做：

MYSQL> SELECT COUNT(foo) FROM database.table WHERE BINARY foo='bar';

BINARY做了魔术!!!

Answer 2

如果要计算foo正好bar的记录，那么您的SQL查询是正确的。

您的PHP代码有问题，您发布的代码根本不起作用（应该计算0条记录）。

Answer 3

MYSQL＆GT; SELECT COUNT（foo）AS FROM FROM database.table WHERE foo ='bar';

$response = mysql_fetch_assoc($surveydata);
echo $response['rows'];

mysql：完全匹配？

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3 个答案: