Question

我有以下PHP代码：

var _eventAggregator = Substitute.For<IEventAggregator>();

我在数据库中获取列名为“name”的所有数据。我怎么能这样输出呢？

 while($row = mysqli_fetch_array($query))
 {
    $data = $row['name'];
 }

Answer 1

你必须将$ data作为数组类型变量。

 while($row = mysqli_fetch_array($query))
 {
   $data[] = $row['name'];
 }

 print_r($data); // required output

Answer 2

 while($row = mysqli_fetch_array($query))
 {
   $data[] = $row['name'];
 }

 print_r($data); // output key wise display like
 Array ( [0] => John [1] => Doe  ) etc.

但按照您的建议输出然后只需添加json_encode()

print_r(json_encode($data)); // output like
["John", "Doe", "Deer"]

Answer 3

$data = [];
while($row = mysqli_fetch_array($query))
{
  $data = $row['name'];
}
echo json_encode($data); // or you can use print_r for debugging.

Answer 4

您需要在数组中使用json_encode（）方法在jquery中接受。重新排列代码如下......

$data = array();
while($row = mysqli_fetch_array($query))
{
   $data[] = $row['name'];
}

$new_array = json_encode($data);
echo $new_array;   // use 'echo' to print. The json_encode() convert $data array to string.

使用PHP输出数组

4 个答案: