无法从PHP上的MySQL响应检索信息
无法从PHP上的MySQL响应检索信息
提问于 2019-05-14 19:28:33
我可以从MySQL服务器获得响应，但似乎不能将其放在变量中
$filmNameList2 = [];
require('connect.php');
$query = "SELECT `title`,`year` FROM `filmList` WHERE year=' (2019)'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
$json_array = array();
while($row=mysqli_fetch_array($result))
{
$json_array[] = $row;
// print_r($row); outputs Array ( [0] => Abruptio [title] => Abruptio [1] => (2019) [year] => (2019) )
}
$filmNameList2[] = $json_array->array[0]->array[0]->title;
// I have tried json_array->array[0]->title; json_array->title;
print_r($filmNameList2);我得到的结果是：
Array ( [0] => )回答 3
Stack Overflow用户
发布于 2019-05-14 19:35:29
$filmNameList2 = array();
require('connect.php');
$query = "SELECT `title`,`year` FROM `filmList` WHERE year=' (2019)'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
while($row=mysqli_fetch_array($result))
{
$object = array(
'title' => $row[0],
'year' => $row[1]
);
array_push($filmNameList2 , $object );
}
echo json_encode($filmNameList2);
print_r($filmNameList2);Stack Overflow用户
发布于 2019-05-14 19:35:55
试试这段代码。
$query = "SELECT `title`,`year` FROM `filmList` WHERE year=' (2019)'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
$json_array = array();
while($row=mysqli_fetch_object($result))
{
$json_array[] = $row;
}
echo json_encode($json_array);
print_r($json_array);Stack Overflow用户
发布于 2019-05-15 03:29:34
$filmNameList2[] = $json_array[0]['title'];
访问第一行，然后访问其中的title元素。
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原文链接：
https://stackoverflow.com/questions/56129363
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