Php 数据不存在时显示零
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"); $numro
for ($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“$x:0
”;
}
这样做将if($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 />";
}