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php - 如何从查询中设置对象属性？

问题描述

我正在尝试将三个值返回到我的前端。我的问题是（或者至少我认为）是它作为布尔值返回。这是我的代码：

<?php
  require "../../inc/dbinfo.inc";

  ini_set("log_errors", 1);
  ini_set("error_log", "/tmp/php-error.log");
  error_log( "#################################################     FT-VENDOR-INVOICE-FILLOUT.PHP    #################################################" );

  $num = $_POST['num'];

  $sql = $conn->prepare("SELECT COUNT(*) AS sq FROM tblVendorInvoices WHERE VendorPOID1 = ?");
  $sql->bind_param("i",$num);
  $sql->execute();
  $hold = $sql->get_result();
  $obj->sq = $hold->fetch_object();
  $obj->sq = round($obj->sq,2);

  $sql2 = $conn->prepare("SELECT POAmount AS poam FROM tblVendorPOs WHERE VENDORPOID = ?");
  $sql2->bind_param("d",$num);
  $sql2->execute();
  $hold2 = $sql2->get_result();
  $obj->poam = $hold2->fetch_object();
  $obj->poam = round($obj->poam,2);

  $sql3 = $conn->prepare("SELECT SUM(VdrInvoiceAmount) AS itd FROM tblVendorInvoices WHERE VendorPOID1 = ?");
  $sql3->bind_param("d",$num);
  $sql3->execute();
  $hold3 = $sql3->get_result();
  $obj->itd = $hold3->fetch_object();
  $obj->itd = round($obj->itd,2);

  echo json_encode($obj);
  exit();
?>

我的错误日志报告：

[01-Aug-2018 09:15:49 America/Toronto] #################################################     FT-VENDOR-INVOICE-FILLOUT.PHP    #################################################
[01-Aug-2018 09:15:49 America/Toronto] PHP Warning:  Creating default object from empty value in /var/www/html/fetch/ft-vendor-invoice-fillout.php on line 14
[01-Aug-2018 09:15:49 America/Toronto] PHP Notice:  Object of class stdClass could not be converted to int in /var/www/html/fetch/ft-vendor-invoice-fillout.php on line 15
[01-Aug-2018 09:15:49 America/Toronto] PHP Notice:  Object of class stdClass could not be converted to int in /var/www/html/fetch/ft-vendor-invoice-fillout.php on line 22
[01-Aug-2018 09:15:49 America/Toronto] PHP Notice:  Object of class stdClass could not be converted to int in /var/www/html/fetch/ft-vendor-invoice-fillout.php on line 29

当前响应：{"sq":1,"poam":1,"itd":1}

期望的响应是这种精确的格式，但数字是错误的。它们是查询值，所以很抱歉我不能给出准确的输出。我尝试过对其中一些行进行类型转换，但没有成功。我真的不明白为什么我的日志中出现这个 int 转换错误，所以对此的解释会很神奇。第一个值应始终为 in，第二个和第三个值应为 float。

另外，我无法理解大多数 PHP 文档的解释，所以请不要参考那些。

标签： phpmysql

解决方案

你误解了什么fetch_object()。它返回一个对象，而不是单个字段。您需要首先引用该字段，然后将其分配给您的对象。

您还可以将两个查询合并为一个，因此下面是为您重构的代码库：

<?php

require '../../inc/dbinfo.inc';

// Retrieve the POST var:
$num = $_POST['num'];

// Create our blank object:
$obj = new stdClass();

// Combines query 1 and 3:
$sql = $conn->prepare('SELECT COUNT(*) AS sq, SUM(VdrInvoiceAmount) AS itd FROM tblVendorInvoices WHERE VendorPOID1 = ?');
$sql->bind_param('i', $num);
$sql->execute();
$hold = $sql->get_result();
$res = $hold->fetch_object();

$obj->sq = round( $res->sq, 2 );
$obj->itd = round( $res->itd, 2 );

$sql2 = $conn->prepare('SELECT POAmount AS poam FROM tblVendorPOs WHERE VENDORPOID = ?');
$sql2->bind_param('d', $num);
$sql2->execute();
$hold = $sql2->get_result();
$res  = $hold->fetch_object();

$obj->pam = round( $res->poam, 2 );

// May also be handy to output JSON headers:
header('Content-type: application/json');
echo json_encode($obj);
exit();

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