PHP数组意外输出

Question

<?php 
$con = mysqli_connect('localhost', 'root', 'root');

if (!$con) {die('Could not connect to databse' . mysqli_error($con));}

mysqli_select_db($con, 'timetable');
mysqli_set_charset($con, 'utf-8');

$sql = "SELECT * FROM time WHERE date = '2019-3-21'";

$result = mysqli_query($con, $sql);
$row = mysqli_fetch_array($result);
$b=array();

for($i = 8; $i<=20; $i+=2){

   if($row[$i.':00'] == 1){
       array_push($b, 1);}

   else{
    array_push($b, 0);}

echo json_encode($b);
}
?>

我期望得到echo 1，1，0，0，1，0，但实际上输出是7个数组，像11,11,1,0,01,1,0,0,0,1，1。我不知道代码出了什么问题。数据库很简单，如图所示。Thx database data

Answer 1

您的循环将多次打印数组，如果您只想在循环结束时打印一次数组，则应该这样做

<?php 
$con = mysqli_connect('localhost', 'root', 'root');

if (!$con) {die('Could not connect to databse' . mysqli_error($con));}

mysqli_select_db($con, 'timetable');
mysqli_set_charset($con, 'utf-8');

$sql = "SELECT * FROM time WHERE date = '2019-3-21'";

$result = mysqli_query($con, $sql);
$row = mysqli_fetch_array($result);
$b=array();

for($i = 8; $i<=20; $i+=2){

   if($row[$i.':00'] == 1){
       array_push($b, 1);}

   else{
    array_push($b, 0);}

}

echo json_encode($b);
?>