我有以下代码，这与其他有效的PHP语句非常相似。但由于某种原因，这个不返回JSON。唯一的区别是这段代码需要整个数据库。
if("method = POST")这不会返回任何内容。但是，如果我放<?php
// Create connection
include_once 'functions.php';
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$state = $_POST["state"];
$db = $state."Standards";
// This SQL statement selects ALL from the table $db
$sql = "SELECT * FROM $db";
// Check if there are results
if ($result = mysqli_query($con, $sql)) {
// If so, then create a results array and a temporary one
// to hold the data
$resultArray = array();
$tempArray = array();
// Loop through each row in the result set
while($row = $result->fetch_object()) {
// Add each row into our results array
$tempArray = $row;
array_push($resultArray, $tempArray);
}
// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);
}
// Close connections
mysqli_close($con);
?>
在echo json_encode($row);命令中，它将返回数据，但当然在每行的单独括号{}中（这里是它的一部分）：
while数据库信息全部适合一个屏幕，所以它并不大。我无法弄清楚为什么将它放入数组的最后一步失败。
答案 0 :(得分：0)
您可以尝试使用// ... code
$resultArray = array();
while($row = $result->fetch_object()) {
// Add each row into our results array
$resultArray[] = $row;
}
// debugging output
print_r($resultArray); // <-- is the result of this output what you expect?
echo json_encode($resultArray);
// ... code
进行调试。这应该会告诉你数组中的内容。
如果import java.util.InputMismatchException;
import java.util.Scanner;
public class Homework1a
{
public static void main(String[] args)
{
int intOne = 0;
int intTwo = 0;
String operation = " ";
int result = 0;
double divResult = 0.0;
// Use the Scanner class to input data
Scanner scannerIn = new Scanner(System.in);
// Display of Program Introduction
System.out.println("This program will take your integer inputs, "
+ "along with the choosen operation and calculate the results");
try
{
// Display for user to input the first integer
System.out.println("Enter the first integer:");
intOne = scannerIn.nextInt();
// Display for user to input the second integer
System.out.println("Enter the second integer:");
intTwo = scannerIn.nextInt();
// Display a list of instructions on the proper call for the operator
System.out.println("");
System.out.println("********** Operator Instructions **********");
System.out.println("\t For addition use +");
System.out.println("\t For subtraction use -");
System.out.println("\t For multiplication use *");
System.out.println("\t For division use /");
System.out.println("\t For modulus use %");
System.out.println("\t For bitwise AND use &");
System.out.println("\t For bitwise inclusive OR use |");
System.out.println("********************************************");
System.out.println("");
// Takes care of the ENTER key negating the String input below
scannerIn.nextLine();
// Display for user to input the chosen operator
System.out.println("Enter the operation:");
operation = scannerIn.nextLine();
}
catch (InputMismatchException ex)
{
System.out.println("Your Integer value is out of range.");
}
}
}
成功json_encoded，则以下代码 应该。但是......让我们看看你的调试会让你发现什么。
<pagination boundary-links="true" total-items="totalItems" ng-model="currentPage" class="pagination-sm" previous-text="‹" next-text="›" first-text="«" last-text="»" items-per-page="1"></pagination>
答案 1 :(得分：0)
当尝试使用编码不正确的文本元素对数组进行编码时，显然json_encode()会无声地失败。我发现here的解决方案是在添加到数组之前对utf8_encode中每个混乱的字符串元素进行编码。
while($row = $result->fetch_object())
{
// Add each row into our results array (and encode in utf8)
$tempArray['grade'] = $row->grade;
$tempArray['category'] = utf8_encode($row->category);
$tempArray['standard'] = utf8_encode($row->standard);
$resultArray[] = $tempArray;
}
// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);