如何读取名称相同但值不同的值

时间:2019-04-18 19:57:51

标签: php android mysql-workbench

我正在编写代码,并且需要从数据库中读取每个零件的数量,问题是所有零件的列数量都是相同的,当我将它们分配给我的Android Studio代码时,为所有请求的字段输出相同的值。我需要如何告诉应用程序获取其他数量的帮助 查询:

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<ul id="list1"></ul>

输出:

<?php
include "connection.php";
$consulta = "SELECT * from parts ;  ";
$result = $con -> query($consulta);

while($fila = $result -> fetch_array()){
    $usuarios[] = array_map('utf8_encode', $fila);
}

echo json_encode($usuarios);
$result -> close();

我想在android studio上显示数量的输出

这是用于从数据库读取的android studio代码:

 [
    {"0":"1","idparts":"1","1":"resistor","partsName":"resistor","2":"13","Quantity":"13"},
    {"0":"2","idparts":"2","1":"capacitor","partsName":"capacitor","2":"10","Quantity":"10"},
    {"0":"3","idparts":"3","1":"inductor","partsName":"inductor","2":"10","Quantity":"10"},
    {"0":"4","idparts":"4","1":"multimeter","partsName":"multimeter","2":"10","Quantity":"10"},
    {"0":"5","idparts":"5","1":"arduino","partsName":"arduino","2":"10","Quantity":"10"},
    {"0":"6","idparts":"6","1":"rasberry","partsName":"rasberry","2":"10","Quantity":"10"},
    {"0":"8","idparts":"8","1":"wires","partsName":"wires","2":"10","Quantity":"10"},
    {"0":"9","idparts":"9","1":"breadboard","partsName":"breadboard","2":"10","Quantity":"10"}
] 

1 个答案:

答案 0 :(得分:0)

fetch_array默认情况下将同时返回一个数字数组和一个assoc数组,您只是感到困惑。

代替我们fetch_assoc,您将只获得assoc数组,这将更有意义

while($fila = $result->fetch_assoc()){
    $usuarios[] = array_map('utf8_encode', $fila);
}
  

同样从您显示的json输出中,除第一行以外,所有其他行中Quantity均为13,而Quantity则为10。