0
$to=$_POST['to'];
$from=$_POST['from'];
$repeat=$_POST['repeat'];
$mysqlquery= mysql_query("select count(patient_id) as idpateint,patient_id from patient where STR_TO_DATE(date_enter,'%d/%m/%Y') between STR_TO_DATE('$repeat','%d/%m/%Y') and STR_TO_DATE('$to','%d/%m/%Y') and patient_type='opd' and patient_id = (select patient_opd from patient where STR_TO_DATE(date_enter,'%d/%m/%Y') between STR_TO_DATE('$from','%d/%m/%Y') and STR_TO_DATE('$to','%d/%m/%Y') and patient_type='opd')")or die('invalid'.mysql_error());
while($mysqlquery1=mysql_fetch_array($mysqlquery)){
echo $mysqlquery1['idpateint'];
}
這是sql查詢。我正在得到如何解決這個問題。請有人幫助解決問題。謝謝如何解決invalidSubquery返回多於1行
還有沒有其他的解決辦法? bcz我得到輸出爲0. – user12345 2014-09-25 17:09:31
也許它是'$ myslquery1 ['idpatient']'?不知道。試試'var_dump($ mysqlquery1)'來看看你真正得到了什麼。 – 2014-09-25 17:11:20
$ mysqlquery = mysql_query(「select count(patient_id)as idpateint,patient_id from STR_TO_DATE(date_enter,'%d /%m /%Y')between STR_TO_DATE('$ repeat','%d /%m /% (從患者中選擇patient_type ='opd'和STR_TO_DATE(date_enter,'%d')和patient_type ='opd'和patient_id in (''從','%d /%m /%Y')和STR_TO_DATE('$到','%d /%m /%Y'))「之間的/%m /%Y'死('invalid'.mysql_error()); ($ mysqlquery1 = mysql_fetch_array($ mysqlquery)){ echo $ mysqlquery1 ['idpateint']; }注意:未定義變量:重複 – user12345 2014-09-26 09:39:48