php - Ajax 成功 jquery php

Question

[{ 
    "SchoolId": "015-08-0034-009-37",
    "SubjectId": "08-0034-00613",
    "StudentId": "T-15981",
    "StudentName": "John"
},{
    "SchoolId": "015-08-0034-009-37",
    "SubjectId": "08-0034-00613",
    "StudentId": "T-15982",
    "StudentName": "Paul"
}]

这是我的 php 页面的 json_encode 响应，当我回显时看起来像这样

Schoolid=015-08-0034-009-37
Subjectid=08-0034-00613
Studentid=T-15981
Studentname=John
Schoolid=015-08-0034-009-37
Subjectid=08-0034-00613
Studentid=T-15982
Studentname=Paul

我可以通过这样做得到这个结果

for (var i = 0; i < data.length; i++) {
    console.log(data[i].schoolid);
    console.log(data[i].subjectid);
    console.log(data[i].studentid);
    console.log(data[i].studentname);
}

我怎样才能使同一个条目的值只生成一次，就像 school id 和 subject id 的值只显示一次一样student id 和 student name 等唯一值将显示两次，因为它有两个不同的值。

我希望得到这样的结果

Schoolid=015-08-0034-009-37
Subjectid=08-0034-00613

Studentid=T-15981
Studentname=John
Studentid=T-15982
Studentname=Paul

在传递 json_encode 之前或在从 php 接收到 json_encode 后在 ajax 中我应该在哪里修复这个？

if ($stmt - > rowCount() > 0) {
    while ($selected_row = $stmt - > fetch(PDO::FETCH_ASSOC)) {
        $basicinfo[] = array('schoolid' => $selected_row['schoolid'], 'subjectid' => $selected_row['subjectid'], 'studentid' => $selected_row['studentid'], 'studentname' => $selected_row['studentname']);
        //$basicinfo1[] = array('schoolid' => $selected_row['schoolid'], 'subjectid' => $selected_row['subjectid']);
        //$basicinfo2[] = array('studentid' => $selected_row['studentid'], 'studentname' => $selected_row['studentname']);
    }
    //$merge = array_merge($basicinfo1 , $basicinfo2);
    //$add = $basicinfo1 + $basicinfo2)
    //$input = array_map("unserialize", array_unique(array_map("serialize", $merge)));
    $input = array_map("unserialize", array_unique(array_map("serialize", $basicinfo)));
    echo json_encode($input, JSON_UNESCAPED_UNICODE);
}

Answer 1

以下代码将帮助您修复 PHP 中的数组

if ($stmt - > rowCount() > 0) {
    //$new_data = array();
    while ($v = $stmt - > fetch(PDO::FETCH_ASSOC)) {
        $tmp_key = $v['SchoolId'].'-'.$v['SubjectId']; //adding a temp key

        $new_data[$tmp_key]['SchoolId'] = $v['SchoolId'];
        $new_data[$tmp_key]['SubjectId'] = $v['SubjectId'];

        $sudky = (isset($new_data[$tmp_key]['Student'])) ? count($new_data[$tmp_key]['Student']) : 0; //getting the key for student child array
        $new_data[$tmp_key]['Student'][$sudky]['StudentId'] = $v['StudentId'];
        $new_data[$tmp_key]['Student'][$sudky]['StudentName'] = $v['StudentName'];
    }
    $new_data = array_values($new_data);
    echo json_encode($new_data, JSON_UNESCAPED_UNICODE);
}

最终的JSON数据如下

[{
    "SchoolId":"015-08-0034-009-37",
    "SubjectId":"08-0034-00613",
    "Student":[
        {"StudentId":"T-15981","StudentName":"John"},
        {"StudentId":"T-15982","StudentName":"Paul"}
    ]
}]

在 Javascript 中，您必须在 for 循环中添加另一个循环，您已经有了它，它会遍历 Student 数据

以下是Javascript部分

<script>
$(document).ready(function($){
    $.ajax({
        type: "post",
        url: "YOURSCRIPT",
        dataType: "json",
        success: function(data){
            if(data!=''){
                for (var i = 0; i < data.length; i++) {
                    alert(data[i].SchoolId);
                    alert(data[i].SubjectId);
                    var student = data[i].Student;
                    for (var j = 0; j < student.length; j++) {
                        alert(student[j].StudentId);
                        alert(student[j].StudentName);
                    }
                }
            }
        }
    });
});
</script>