Question

我使用ajax将数据发送到查询mysql并返回结果的php页面。我在我的页面中有一个包含dept值的变量。但是，如果我var_dump（$ _ POST），我发现var是一个数组。

如果我在查询中手动输入值，则返回数据。但是，只是没有使用我的$ dept var。

如何解码此数组以使变量可用于我的查询。感谢

ajax代码

dept=DEMOBILL

在firebug标签中发帖

<?php

    $dept = $_POST['dept']; <--- array?
    //open connection to mysql db
    $connection = mysqli_connect("localhost","root","","sample") or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select custref from boxes where department = '".$dept."' and status = 1";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray = array();
    while($row = mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }
    echo json_encode($emparray);

    //close the db connection
    mysqli_close($connection);
?>

deptdata.php

doctrine:
    orm:
        entity_managers:
            default:
                mappings:
                    ApplicationSonataMediaBundle: ~
                    SonataMediaBundle: ~


    dbal:
        types:
            json: Sonata\Doctrine\Types\JsonType

sonata_media:
    # if you don't use default namespace configuration
    #class:
    #    media: MyVendor\MediaBundle\Entity\Media
    #    gallery: MyVendor\MediaBundle\Entity\Gallery
    #    gallery_has_media: MyVendor\MediaBundle\Entity\GalleryHasMedia
    db_driver: doctrine_orm # or doctrine_mongodb, doctrine_phpcr it is mandatory to choose one here
    default_context: default # you need to set a context
    contexts:
        default:  # the default context is mandatory
            providers:
                - sonata.media.provider.dailymotion
                - sonata.media.provider.youtube
                - sonata.media.provider.image
                - sonata.media.provider.file
                - sonata.media.provider.vimeo

            formats:
                small: { width: 100 , quality: 70}
                big:   { width: 500 , quality: 70}

    cdn:
        server:
            path: /uploads/media # http://media.sonata-project.org/

    filesystem:
        local:
            directory:  "%kernel.root_dir%/../web/uploads/media"
            create:     false

Answer 1

$myArray = json_decode($data, true);

echo $myArray[0]['id']; // Fetches the first ID
echo $myArray[0]['c_name']; // Fetches the first c_name
// ...
echo $myArray[2]['id']; // Fetches the third ID
// etc..

如果没有将true作为第二个参数传递给json_decode，它会将其作为对象返回：

echo $myArray[0]->id;

我如何解码json数组

1 个答案: