Php 数据不存在时显示零

Php 数据不存在时显示零,php,mysql,Php,Mysql,如果数据不存在,我想显示0,如示例中所示。我尝试过使用numrows,但仍然不起作用…很抱歉代码不好。我还在学习。任何意见都非常感谢 for ($x = 2013; $x<=2017; $x++) { $result= $myDB->query("SELECT * FROM ".$myDB->prefix("mydata")." WHERE year='$x' AND data_id=8"); $numro

如果数据不存在,我想显示0,如示例中所示。我尝试过使用numrows,但仍然不起作用…很抱歉代码不好。我还在学习。任何意见都非常感谢

    for ($x = 2013; $x<=2017; $x++) 
    { 

        $result= $myDB->query("SELECT * FROM ".$myDB->prefix("mydata")." WHERE year='$x' AND data_id=8");
                 $numrows = $myDB->getRowsNum($result);
                  while($row = $myDB->fetchArray($result))
                   {

                   $year=$row['year']; 
                  $total=$row['total']; 
                   echo "$year : $total <br />"; 
                   } 

     }
期望输出

    2013 : 456 
    2014 : 103
    2015 : 0
    2016 : 45 
    2017 : 34 
我试过这样的东西,但它不起作用

for ($x = 2013; $x<=2017; $x++) { 

$result= $myDB->query("SELECT * FROM ".$myDB->prefix("mydata")." WHERE year='$x' AND data_id=8");
 $numrows = $myDB->getRowsNum($result);
 while($row = $myDB->fetchArray($result))
     { 

       if ($numrows > 0) {
           $year=$row['year']; 
       $total=$row['total']; 
    echo "$year : $total <br />"; 
       }
    else
    {
        echo "$x : 0 <br />"; 
    }


    } 
    }
for($x=2013;$xquery(“从“$myDB->prefix(“mydata”)中选择*”,其中年份=“$x”,数据id=8”);
$numrows=$myDB->getRowsNum($result);
而($row=$myDB->fetchArray($result))
{ 
如果($numrows>0){
$year=$row['year'];
$total=$row['total'];
echo“$year:$total
”; } 其他的 { 回声“$x:0
”; } } }
如果($numrows>0){
而($row=$myDB->fetchArray($result))
{ 
$year=$row['year'];
$total=$row['total'];
echo“$year:$total
”; } } 其他的 { 回声“$x:0
”; }
这样做

if($numRows>0)
移动到
循环之外
for ($x = 2013; $x<=2017; $x++) { 

$result= $myDB->query("SELECT * FROM ".$myDB->prefix("mydata")." WHERE year='$x' AND data_id=8");
 $numrows = $myDB->getRowsNum($result);
 while($row = $myDB->fetchArray($result))
     { 

       if ($numrows > 0) {
           $year=$row['year']; 
       $total=$row['total']; 
    echo "$year : $total <br />"; 
       }
    else
    {
        echo "$x : 0 <br />"; 
    }


    } 
    }
    if ($numrows > 0) {
 while($row = $myDB->fetchArray($result))
     { 
       $year=$row['year']; 
       $total=$row['total']; 
    echo "$year : $total <br />"; 
     }
}
    else
    {
        echo "$x : 0 <br />"; 
    }