未定義的索引最大功能不工作在PHP

Question

我面臨着我的PHP代碼的問題。我已經嘗試了堆棧溢出的建議，但該代碼無法正常工作。這裏我使用的是max和min的功能。還嘗試功能mysql_fetch_rows()和mysql_fetch_assoc()。它向我顯示錯誤undefined index:facilities_total in line...我也在下面的代碼中指出我的錯誤。請建議如何解決此問題。

這是我的PHP與MySQL查詢

$sqlMax= "SELECT projects_detail.Project_name,max(facility_rating.facilities_total) 
    AS MAX_FACILITIES_TOTAL 
    FROM facility_rating LEFT OUTER JOIN projects_detail 
    on facility_rating.project_id = projects_detail.project_id 
    group by facility_rating.facilities_total DESC LIMIT 1"; 
$resultMax=mysql_query($sqlMax); 
echo "<table border cellpadding=3>"; 
while($rowsMax=mysql_fetch_array($resultMax)){ 
    echo "<tr>"; 
    echo "<td>".$rowsMax['Project_name'] . "</td> "; 
    echo "<td>".$rowsMax['facilities_total'] . "</td> "; //error undefined index:facilities_total 
    echo "</tr>"; 
} 
echo "</table>"; 

Answer 1

試試這個：

echo "<td>".$rowsMax['MAX_FACILITIES_TOTAL'] . "</td> "; 

而不是

echo "<td>".$rowsMax['facilities_total'] . "</td> "; 

在SQL您正在使用別名facilities_total，所以行取自字段名稱MAX_FACILITIES_TOTAL而不是facilities_total。

Answer 2

你的SQL命名輸出列在這裏：

AS MAX_FACILITIES_TOTAL 

所以，你想要的PHP變量會

$rowsMax['MAX_FACILITIES_TOTAL'] 

Answer 3

可以看出你是給別名在您的查詢，您使用的列名您的php代碼：

$sqlMax= "SELECT projects_detail.Project_name,max(facility_rating.facilities_total) 
    AS MAX_FACILITIES_TOTAL 
    FROM facility_rating LEFT OUTER JOIN projects_detail 
    on facility_rating.project_id = projects_detail.project_id 
    group by facility_rating.facilities_total DESC LIMIT 1"; 
$resultMax=mysql_query($sqlMax); 
echo "<table border cellpadding=3>"; 
while($rowsMax=mysql_fetch_array($resultMax)){ 
    echo "<tr>"; 
    echo "<td>".$rowsMax['Project_name'] . "</td> "; 
    // Fix: Changed 'facilities_total' to 'MAX_FACILITIES_TOTAL' below. 
    echo "<td>".$rowsMax['MAX_FACILITIES_TOTAL'] . "</td> "; //error undefined index:facilities_total 
    echo "</tr>"; 
} 
echo "</table>";