Question

我有数据库查询，它返回一个字符串值数组。我希望它们以逗号分隔并附加到另一个字符串。

$sql1 = "select location.name as name from location,destination where destination.name='".$destination."' AND location.destination_id=destination.destination_id";
$result1 = $conn->query($sql1);
if ($result1->num_rows > 0) {
  $insert = "";
    while($locations = $result1->fetch_assoc()) {     
  foreach ($locations as $location) {
$insert .= $location['name'] . ",";
      }
   }

$insert = rtrim($insert, ",");

 }

$sql2 = "select destination.destination_description as description from destination where destination.name='".$destination."'";
$result2 = $conn->query($sql2);
if ($result2->num_rows > 0) {
    while($row = $result2->fetch_assoc()) {
    $response->speech = $destination." is the ".$row["description"].". ".$insert . "are some of the places you can try.";
   echo json_encode($response);
  }
}

此$ insert数组未提供预期结果。它只给出“N”。

Answer 1

您可以删除foreach行（$ locations as $ location）

您已经将所有行提取到位置，因此您只需附加位置['name']。您的foreach已检索字符串$ locations

中的每个字符

Answer 2

我认为你应该看看可能对你有帮助的爆炸功能Explode function in php

Answer 3

如果我理解你的问题。你需要使用php的implode（）函数。试试这个： -

if ($result1->num_rows > 0) {
$ImplodeArray =array();
while($locations = $result1->fetch_assoc()) {     
    foreach ($locations as $location) {
        $ImplodeArray[] = $location['name'];
    }
}
$insert = implode(',',$ImplodeArray);
echo $insert;
}

希望它有所帮助！

使用字符串附加php数组值不会产生预期的输出

3 个答案: