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php - 无法从 PHP 上的 MySQL 响应中检索信息

转载 作者:行者123 更新时间:2023-11-29 16:01:48 25 4
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我能够从 MySQL 服务器获取响应,但我似乎无法将其放入变量中

$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:

Array ( [0] => )

最佳答案

$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);

关于php - 无法从 PHP 上的 MySQL 响应中检索信息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56129363/

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