Question

我以JSON格式从MySQL数据库中获取一些数据，但我有一个奇怪的问题。

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，但是如果我写的话：

<?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。都没有错误代码......

表格如下：

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＆＃34; Tienda＆＃34;它不起作用。或者＆＃34; Usuario＆＃34;行。如果我只包含＆＃34; id＆＃34;排它很好......

Answer 1

我修改了它改变了编码......

<?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");并修改相对于JSON

编码的最后一行

从MySQL获取数据，查询不同的问题

1 个答案: