php - MySQL 的 JSON 输出有效，但随后出现白屏？

Question

我使用以下代码从 MySQL 表生成 JSON。当我只生成 2 个数组时效果很好，但由于某种原因，当我尝试生成 3 或 4 个数组时，我得到了“死亡白屏”。我认为它正在发生，因为我需要提高我的 PHP 内存限制，但我已经这样做了，但仍然遇到同样的问题。见下文:

有效的代码是:

<?php
    //Create Database connection
    $db = mysql_connect("localhost","user","dbpassword");
    if (!$db) {
        die('Could not connect to db: ' . mysql_error());
    }

    //Select the Database
    mysql_select_db("dbname",$db);

    //Replace * in the query with the column names.
    $result = mysql_query("select * from customer", $db);  

    //Create an array
    $json_response = array();

    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        $row_array['customerfname'] = $row['customerfname'];
        $row_array['customerlname'] = $row['customerlname'];

        //push the values in the array
        array_push($json_response,$row_array);
    }
    echo json_encode($json_response);

    //Close the database connection
    fclose($db);

?>

但是当我尝试调用更多值时，这是我得到空白屏幕的地方:

<?php
    //Create Database connection
    $db = mysql_connect("localhost","user","dbpassword");
    if (!$db) {
        die('Could not connect to db: ' . mysql_error());
    }

    //Select the Database
    mysql_select_db("dbname",$db);

    //Replace * in the query with the column names.
    $result = mysql_query("select * from customer", $db);  

    //Create an array
    $json_response = array();

    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        $row_array['customerfname'] = $row['customerfname'];
        $row_array['customerlname'] = $row['customerlname'];
        $row_array['customeremail'] = $row['customeremail'];
        $row_array['customertel'] = $row['customertel'];

        //push the values in the array
        array_push($json_response,$row_array);
    }

   var_dump($json_response);

   echo json_encode($json_response);



?>

更新**

错误似乎是我的最后一行 (echo json_encode ($json_response);

当我在最后一行之前添加 print_r($row) 时，会生成以下输出。

Array ( [id] => 1 [customerfname] => First [customerlname] => Last [customeremail] => test@test.com [customerphone] => 000-000-0000

更新 2

在 echo json_encode 行之前添加 var_dump($json_response); 会导致空白屏幕。

Answer 1

供日后引用:如果其他人因为有类似问题访问此问题，则出现空白页错误，因为查询返回了太多行和json_encode() 似乎无法处理它。

---

尝试以下方法，并与我们分享您的结果，

<?php
    ini_set('display_errors', 1); 
    error_reporting(E_ALL);

    //Create Database connection
    $db = mysql_connect("localhost","user","dbpassword");
    if (!$db)
        die('Could not connect to db: ' . mysql_error());

    mysql_select_db("dbname",$db);

    mysql_query('SET CHARACTER SET utf8');

    $result = mysql_query("select `customerfname`, `customerlname`, `customeremail`, `customertel`  from customer", $db) or die("Error: ".mysql_error());  

    $json_response = array();

    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        $json_response[] = $row; 
    }

   var_dump($json_response);

    /* If the var_dump works, you can try and uncomment this section
    if (function_exists('json_encode')) 
    {
        echo json_encode($json_response);
        echo "JSON Error: ".json_last_error();
    }
    else { echo "json_encode() is not supported"; }
    */
?>