SyntaxError：" JSON.parse：unexpected non-whitespace ..."从PHP返回JSON时 [英] SyntaxError: "JSON.parse: unexpected non-whitespace ..." when returning JSON from PHP

问题描述

我遇到一个问题，从PHP查询返回的JSON无效，我不确定为什么;我还在学习。当排除数据类型时，以下代码返回：

I'm running into an issue where the JSON returned from a PHP query is not valid and I'm not really sure why; I'm still learning. When the datatype is excluded the below code returns:

{"Customer_ID":"0", "FirstName":"John", "LastName":"Smith"}
{"Customer_ID":"1", "FirstName":"Jane", "LastName":"Smith"}

否则返回：

SyntaxError: "JSON.parse: unexpected non-whitespace character after ..."

我认为这可能是因为记录没有在单个JSON响应中返回，但由于并发响应是JSON，我无法看到这个问题。有任何想法吗？有什么建议？随意指出语义问题。

I thought this might be because the record is not being returned in a single JSON response, but I can't see that being the issue as the concurrent responses are JSON. Any ideas? Any suggests? Feel free to point out the semantic issues.

HTML：

getRecord("*", "customer", "");

JavaScript：

function getRecord(field, table, condition) {
    var request = $.ajax({
        url: "./handler.php",
        method: "GET",
        dataType: "JSON",
        cache: "false",
        data: {
            action: "SELECT",
            field: `${field}`,
            table: `${table}`,
            condition: `${condition}`,
        },
    });

    request.done(function(data, status, xhr) {
        console.log(data, status, xhr);
    });

    request.fail(function(xhr, status, error) {
        console.log(xhr, status, error);
    });

};

PHP：

<?php

    # IMPORT SETTINGS.
    include "settings.php";

    # FUNCTION DISPATCHER.
    switch($_REQUEST["action"]) {

        case "SELECT":
            getRecord($conn);
            break;

        default:
            printf('Connection Error: Access Denied.');
            mysqli_close($conn);
    }

    # LIST OF COLUMNS THAT WE NEED.

    function getRecord($conn) {
        $table = $_REQUEST["table"];
        $field = $_REQUEST["field"];
        $condition = $_REQUEST["condition"];

        if (!empty($condition)) {
            $query = "SELECT $field FROM $table WHERE $condition";
        } else {
            $query = "SELECT $field FROM $table";
        }

        if ($result = mysqli_query($conn, $query)) {
            while ($record = mysqli_fetch_assoc($result)) {
                echo json_encode($record);
            }
        }

        # CLOSE THE CONNECTION.
        mysqli_close($conn);

    }

?>

推荐答案

您的JSON无效，因为它包含多个对象。您需要做的是将所有结果放入一个数组中，然后回显 json_encode 。尝试这样的事情：

Your JSON is not valid because it consists of multiple objects. What you need to do is put all your results into an array and then echo the json_encode of that. Try something like this:

    $records = array();
    if ($result = mysqli_query($conn, $query)) {
        while ($records[] = mysqli_fetch_assoc($result)) {
        }
    }
    echo json_encode($records);

这将为您提供如下所示的输出：

This will give you an output that looks something like this:

[
    {"Customer_ID":"0", "FirstName":"John", "LastName":"Smith"},
    {"Customer_ID":"1", "FirstName":"Jane", "LastName":"Smith"}
]

您可以通过以下方式访问Javascript中的每个元素

and you can access each of the elements in your Javascript by something like

let customer = data[0].FirstName + ' ' + data[0].LastName;

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