我的数据库中有多个表，我想将它们连接在一起，为 GET 请求创建一个嵌套的 JSON。有一次我想根据数据库中的值添加一个数组，例如(类型 === "fixValue")。
... $sql = "SELECT name, type
FROM section INNER JOIN question ON section.section_id = question.section_id
WHERE question.section_id = ".$row_section['section_id']."";
$stmtq = $db->query($sqlq);
$stmtq -> execute();
while( $row_question = $stmtq->fetch(PDO::FETCH_ASSOC)){
$question_array['name'] = $row_question['name'];
$question_array['type'] = $row_question['type'];
if($row_question['type'] == "fixValue"){
$question_array['options'] = array();
}
array_push($section_array['sectionQuestion'], $question_array);
} ....
我现在的 json 输出是:
{
"name": "Test",
"something": "Test",
"array1": [
{
"name": "Section",
"array2": [
{
"name": "Something",
"type": "fixValue",
"options": []
},
{
"name": "Sometthing2",
"type": "notFixValue",
"options": []
}
]
}
]}
我想要的输出:
{
"name": "Test",
"something": "Test",
"array1": [
{
"name": "Section",
"array2": [
{
"name": "Something",
"type": "fixValue",
"options": []
},
{
"name": "Sometthing2",
"type": "notFixValue",
}
]
}
]}
因此，选项数组不仅会添加到值为“fixValue”的项目中。类型值与“fixValue”不同的项目应该没有“options”数组。 有什么想法可以实现这样的目标吗？
最佳答案
尝试使用计数器代替 array_push:
... $sql = "SELECT name, type
FROM section INNER JOIN question ON section.section_id = question.section_id
WHERE question.section_id = ".$row_section['section_id']."";
$stmtq = $db->query($sqlq);
$stmtq -> execute();
$cnt = 0;
while( $row_question = $stmtq->fetch(PDO::FETCH_ASSOC)){
$question_array['name'] = $row_question['name'];
$question_array['type'] = $row_question['type'];
if($row_question['type'] == "fixValue"){
$question_array['options'] = array();
}
$section_array['sectionQuestion'][$cnt] = $question_array;
unset($question_array);
$cnt++;
} ....
关于php - 如果数据库中的值等于字符串，则将数组添加到嵌套的 JSON，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53662397/