Question

这段代码json_encode行没有回显

    include "connect.php";
$sql="select * from customers"; 


$arr = array();
$result=mysqli_query($conn,$sql);
while ($row = mysqli_fetch_assoc($result))
{
    $firstname = $row['firstname']; 
    $lastname = $row['lastname'];
    $arr[] = array('lastname' => $lastname, 'lastname' => $lastname);

}

echo json_encode($arr);// results in nothing.

我做错了什么？请帮忙

Answer 1

即使您在空数组上执行echo json_encode($arr);，也应该看到结果[]。所以问题可能在mysqli_代码中的某个地方。

请检查通过任何mysql数据库扩展所做的所有调用的状态。通过这种方式，当您出现某种错误时会通知您。

include "connect.php";

// this should probably go in the 'connect.php'
// this is just an example no a difinitive piece of code
function db_err(&$conn, &$sql, $lineNo)
{
    if ( $conn_errno($conn) ) {
        echo sprinf( 'Query %s failed on line %d Code:%d Message:%s', 
                      $sql, 
                      $lineNo,
                      mylsqi_errno($conn), 
                      mylsqi_error($conn));
        exit;
    }
}

$sql="select * from customers"; 

$result = mysqli_query($conn,$sql);
db_err($conn,$sql, __LINE__);

$arr = array();
while ($row = mysqli_fetch_assoc($result)) {
    $firstname = $row['firstname']; 
    $lastname = $row['lastname'];
    $arr[] = array('lastname' => $firstname, 'lastname' => $lastname);

}

echo json_encode($arr);

json_encode无法在PHP中运行

1 个答案: