php - 从MySQL获取数据，不同查询的问题

Question

我正在从 MySQL 数据库中以 JSON 格式获取一些数据，但遇到了一个奇怪的问题。

如果我写代码:

<?php

    //Step1 - Access Database
    ini_set( "display_errors", "1" );
    $db = mysqli_connect('-','-','-','-')
    or exit(mysqli_connect_error());

    //Step2 - Create query and execute it

    $query = "SELECT `id` FROM `Tiendas`";
    $result = mysqli_query($db, $query) or exit(mysqli_error($db));

    //Step 3 - Configure results

    $data = array();
    while ( $row = $result->fetch_assoc() ) {
      $data[] = $row;
    }

    //Step 4 - Close connection and send results
    header( 'Content-Type: application/json' );
    mysqli_close($db);
    echo json_encode($data,JSON_UNESCAPED_UNICODE);
?>

效果很好，但如果我写:

<?php

    //Step1 - Access Database
    ini_set( "display_errors", "1" );
    $db = mysqli_connect('-','-','-','-')
    or exit(mysqli_connect_error());

    //Step2 - Create query and execute it

    $query = "SELECT `Tienda` FROM `Tiendas`";
    $result = mysqli_query($db, $query) or exit(mysqli_error($db));

    //Step 3 - Configure results

    $data = array();
    while ( $row = $result->fetch_assoc() ) {
      $data[] = $row;
    }

    //Step 4 - Close connection and send results
    header( 'Content-Type: application/json' );
    mysqli_close($db);
    echo json_encode($data,JSON_UNESCAPED_UNICODE);
?>

它没有显示任何内容。也没有错误代码...

表格是这样的:

CREATE TABLE `Tiendas` (
    `id` INT(11) NOT NULL AUTO_INCREMENT,
    `Usuario` VARCHAR(50) NULL DEFAULT NULL,
    `Tienda` VARCHAR(50) NULL DEFAULT NULL,
    PRIMARY KEY (`id`)
)
COLLATE='utf8_general_ci'
ENGINE=InnoDB
AUTO_INCREMENT=217
;

如果我在 SELECT 语句中包含“Tienda”或“Usuario”行，则不起作用。如果我只包含“id”行，效果很好......

Answer 1

我通过更改编码修复了它...

<?php

    //Step1 - Access Database
    ini_set( "display_errors", "1" );
    $db = mysqli_connect('localhost','fgallegof','qU5v6f5$','cdm')
    or exit(mysqli_connect_error());
    mysqli_set_charset($db, "utf8");

    //Step2 - Create query and execute it

    $query = "SELECT `Tienda` FROM `Tiendas`";
    $result = mysqli_query($db, $query) or exit(mysqli_error($db));

    //Step 3 - Configure results

    $data = array();
    while ( $row = $result->fetch_assoc() ) {
      $data[] = $row;
    }

    //var_dump($data);

    //Step 4 - Close connection and send results
    header( 'Content-Type: application/json' );
    mysqli_close($db);
    echo json_encode($data);
?>

我需要添加mysqli_set_charset($db, "utf8");并修改最后一行相对于JSON的编码