dpwle46882 2015-09-19 20:10
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json_encode没有返回任何内容

I am trying to convert my MYSQL table data into JSON. I am trying with json_encode(). But it does not work. It does not return anything. I have checked the console,doesn't even throw any errors. What am i missing?

<?php
    //open connection to mysql db
    $connection = mysqli_connect("localhost","root","","maps") or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select * from locations";
    $result = mysqli_query($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);
?>
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5条回答 默认 最新

  • douqiao2008 2015-09-19 20:43
    关注

    try this

    while ( $row = $result->fetch_assoc() ){
        $emparray[] = json_encode($row);
    }
    echo json_encode( $emparray );
    

    or

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

    or

    $emparray = $result->fetch_all( MYSQLI_ASSOC );
    echo json_encode( $emparray );
    

    instead of

     while($row =mysqli_fetch_assoc($result))
        {
            $emparray[] = $row;
        }
        echo json_encode($emparray);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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