javascript - js取得php的回傳結果問題

<code> $.ajax({
            url: '/ajax.php',
            type: 'POST',
            contentType: 'application/json; charset=UTF-8',
            crossDomain: true,
            dataType: 'json',
            data: JSON.stringify(data),
            success: function(response) {
            
                alert(response);
                console.log(response);

                $("#spinny").hide();

                var data = response.hits.hits;
                console.log(data);
                var source = null;

                if (data.length > 0) {
                    $("#resultsHeader").html(data.length + " Results").show();
                    for (var i = 0; i < data.length; i++) {
                        source = data[i].fields;
                    }

                } else {
                  //$("#resultsHeader").html("No Results").show();
                    showErrorMessage("#error-container", "<strong>Ooops!</strong> No results found! Please try again.","alert-danger", true, 3000);
                }

            },
            error: function(jqXHR, textStatus, errorThrown) {
                var jso = jQuery.parseJSON(jqXHR.responseText);
                error_note('section', 'error', '(' + jqXHR.status + ') ' + errorThrown + ' --<br />' + jso.error);
            }
            </code>

以上的js的提交過程。

以下是ajax.php的程式碼

<code>

<?php
    
    require_once('bootstrap.php');
    
    $url                   = ELASTICSEARCH_URL.'/test/_search';
    $content               = file_get_contents('php://input');
    $ch                    = curl_init($url);
    curl_setopt($ch,       CURLOPT_CUSTOMREQUEST, "POST");
    curl_setopt($ch,       CURLOPT_POSTFIELDS, $content);
    curl_setopt($ch,       CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch,       CURLOPT_CONNECTTIMEOUT, 0);
    curl_setopt($ch,       CURLOPT_TIMEOUT, 5000);
    curl_setopt($ch,       CURLOPT_HTTPHEADER, array('Content-Type: application/json', 'Content-Length: ' . strlen($content)));
    
    $result= curl_exec($ch);
    $arr = json_decode($result);
    if ($arr->hits->total>0) {
        foreach ($arr->hits->hits as $es) {
            $source = $es->_source;
            foreach ($source as $key => $value) {
                echo "$key".":"."$value";
            }
        }
    }
    ?>
                
                </code>

問題:js可以正常產生請求到php，但是取不到php的回傳值，不管是 $result還是$key,請問這個怎麼回事？
看js的報錯，會得到如下的一個提示

<code>"VM2900:1 Uncaught SyntaxError: Unexpected token y in JSON at position 1"
            </code>

回覆內容：

<code> $.ajax({
            url: '/ajax.php',
            type: 'POST',
            contentType: 'application/json; charset=UTF-8',
            crossDomain: true,
            dataType: 'json',
            data: JSON.stringify(data),
            success: function(response) {
            
                alert(response);
                console.log(response);

                $("#spinny").hide();

                var data = response.hits.hits;
                console.log(data);
                var source = null;

                if (data.length > 0) {
                    $("#resultsHeader").html(data.length + " Results").show();
                    for (var i = 0; i < data.length; i++) {
                        source = data[i].fields;
                    }

                } else {
                  //$("#resultsHeader").html("No Results").show();
                    showErrorMessage("#error-container", "<strong>Ooops!</strong> No results found! Please try again.","alert-danger", true, 3000);
                }

            },
            error: function(jqXHR, textStatus, errorThrown) {
                var jso = jQuery.parseJSON(jqXHR.responseText);
                error_note('section', 'error', '(' + jqXHR.status + ') ' + errorThrown + ' --<br />' + jso.error);
            }
            </code>

以上的js的提交過程。

以下是ajax.php的程式碼

<code>

<?php
    
    require_once('bootstrap.php');
    
    $url                   = ELASTICSEARCH_URL.'/test/_search';
    $content               = file_get_contents('php://input');
    $ch                    = curl_init($url);
    curl_setopt($ch,       CURLOPT_CUSTOMREQUEST, "POST");
    curl_setopt($ch,       CURLOPT_POSTFIELDS, $content);
    curl_setopt($ch,       CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch,       CURLOPT_CONNECTTIMEOUT, 0);
    curl_setopt($ch,       CURLOPT_TIMEOUT, 5000);
    curl_setopt($ch,       CURLOPT_HTTPHEADER, array('Content-Type: application/json', 'Content-Length: ' . strlen($content)));
    
    $result= curl_exec($ch);
    $arr = json_decode($result);
    if ($arr->hits->total>0) {
        foreach ($arr->hits->hits as $es) {
            $source = $es->_source;
            foreach ($source as $key => $value) {
                echo "$key".":"."$value";
            }
        }
    }
    ?>
                
                </code>

問題:js可以正常產生請求到php，但是取不到php的回傳值，不管是 $result還是$key,請問這個怎麼回事？
看js的報錯，會得到如下的一個提示

<code>"VM2900:1 Uncaught SyntaxError: Unexpected token y in JSON at position 1"
            </code>

php回傳的值不是json格式的，js沒有辦法解析

echo json_encode()