Question

问题我通过选择每个数据集并逐行输出来输出一组数据。

问题是否有一种获取此数据并将其存储为数组的方法，然后我可以将其转换为字符串值集合？

$sql = "SELECT * FROM respondent_data WHERE respondent_firstname = 'John'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {

  while($row = mysqli_fetch_assoc($result)) {

    $categoriesTest[] =   $row["respondent_sdo"];
    $row["respondent_dcto"];
    $row["respondent_ed"];
    $row["respondent_ca"];
    $row["respondent_dhpt"];
    $row["respondent_irt"];
    $row["respondent_gl"];
    $row["respondent_il"];      

      // Turn my output into an array ready to be used for the JSON string

  }

}

示例

所以这些值中的每一个都从我需要它们的列行输出一个整数，以便变成一个数组，如：2,4,3,5 ....

Answer 1

这应该可以解决问题。

$sql = "SELECT * FROM respondent_data WHERE respondent_firstname = 'John'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {

  $i=0;
  while($row = mysqli_fetch_assoc($result)) {

    $categoriesTest[$i]["respondent_sdo"] = $row["respondent_sdo"];
    $categoriesTest[$i]["respondent_dcto"] = $row["respondent_dcto"];
    $categoriesTest[$i]["respondent_ed"] = $row["respondent_ed"];
    $categoriesTest[$i]["respondent_ca"] = $row["respondent_ca"];
    $categoriesTest[$i]["respondent_dhpt"] = $row["respondent_dhpt"];
    $categoriesTest[$i]["respondent_irt"] = $row["respondent_irt"];
    $categoriesTest[$i]["respondent_gl"] = $row["respondent_gl"];
    $categoriesTest[$i]["respondent_il"] = $row["respondent_il"];      

    $i++;

  }

  var_dump($categoriesTest);

}

Answer 2

如果您只想要从MySQL调用中查询的所有列中的特定关联属性，只需将它们设置在具有各自属性的数组中：

$categoriesTest = array();
$sql    = "SELECT * FROM `respondent_data` WHERE `respondent_firstname` = 'John'";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
    $categoriesTest[] = array(
        'respondent_sdo'    => $row['respondent_sdo'],
        'respondent_dcto'   => $row['respondent_dcto'],
        'respondent_ed'     => $row['respondent_ed'],
        'respondent_ca'     => $row['respondent_ca'],
        'respondent_dhpt'   => $row['respondent_dhpt'],
        'respondent_irt'    => $row['respondent_irt'],
        'respondent_gl'     => $row['respondent_gl'],
        'respondent_il'     => $row['respondent_il']
    );
}
$categoriesTest = json_encode($categoriesTest); // get JSON

但是，如果您想要查询所有列，但要创建自定义元素，则您需要使用array_merge：

$categoriesTest = array();
$sql    = "SELECT * FROM `respondent_data` WHERE `respondent_firstname` = 'John'";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
    $categoriesTest[] = array_merge($row, array(
        'custom_value_1'    => 'test',
        'custom_value_2'    => 'test2'
    ));
}
$categoriesTest = json_encode($categoriesTest); // get JSON

将SQL行转换为数组

2 个答案: