我创建了一个不返回任何值的函数:
$j2_2= mysqli_query($con, "select * from Hagrala where Ale = 7 order by Mispar desc limit 1;");
$j2_3 = mysqli_fetch_array($j2_2);
$spade_7_2=$j2_3['Mispar'];
这有效
$v='9';
$j2_2= mysqli_query($con, "select * from Hagrala where Ale = $v order by Mispar desc limit 1;");
$j2_3 = mysqli_fetch_array($j2_2);
$spade_9_2=$j2_3['Mispar'];
这也有效 但是使用函数foo它不起作用:
function foo($arg_1)
{
$t_1 = mysqli_query($con, "select * from Hagrala where Ale = $arg_1 order by Mispar desc limit 1;");
$t_2 = mysqli_fetch_array($t_1);
return $t_2;
}
$spade_10_2=foo('10');
有什么想法吗?
答案 0 :(得分:2)
$con
。 $con
位于全局范围内,除非您通过可用方式实现,否则您的函数无法使用function foo($arg_1, $con) {
$t_1 = mysqli_query($con, "select * from Hagrala where Ale = $arg_1 order by Mispar desc limit 1;");
$t_2 = mysqli_fetch_array($t_1);
return $t_2;
}
$spade_10_2=foo('10', $con);
。执行此操作的最佳方法是将其作为参数传递给函数。
{{1}}
答案 1 :(得分:0)
在函数参数中传递$con
以在函数中访问它。
function foo($arg_1,$con)
{
$t_1 = mysqli_query($con, "select * from Hagrala where Ale = $arg_1 order by Mispar desc limit 1;");
$t_2 = mysqli_fetch_array($t_1);
return $t_2;
}
$spade_10_2=foo('10');