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我很困惑为什么我的 php mysql 函数不能在远程服务器上运行，而它在我的本地主机上运行良好。

<?php
function getAllMachines(){



    if(!isset($_GET['cat'])){
   global $db_con;

    $getmachine_qry = "SELECT * FROM MACHINES";

    $run_getmachine_qry = mysqli_query($db_con, $getmachine_qry);

    while($row_getmachine = mysqli_fetch_array($run_getmachine_qry)){

        $machine_ID = $row_getmachine['id'];
        $machine_cat = $row_getmachine['fk_cat_id'];
        $machine_title = $row_getmachine['machine_title'];
        $machine_year = $row_getmachine['year'];
        $machine_image = $row_getmachine['mac_image']; 
        $machine_price = $row_getmachine['price'];


    $getCats_qry = "SELECT * FROM categories where id=$machine_cat";

    $run_getCats = mysqli_query($db_con, $getCats_qry);

    while($row_getCats = mysqli_fetch_array($run_getCats)){

        $cat_title = $row_getCats['cat_title'];

    }

        echo "<div class='clearfix' style='float: left; margin: 20px; padding: 10px; box-shadow: 0px 0px 4px 2px rgba(0,0,0,0.4);'>"
                . "<a href='detail.php?mac_id=$machine_ID'><img class='img-thumbnail' src='images/$machine_title/$machine_image' width='250px' /></a>"
                . "<h4><b>$machine_title</b></h4>"
                . "<h5 class='text-danger'>$cat_title</h5>"
                . "<p style='float:left;'>Year: <b class='text-warning'>$machine_year</b></p><br />"
                . "<h4 class='bg-success text-white'><b>&pound; $machine_price</b></h4>"
                . "</div>";



 }
 }

}
?>
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1 回答 1

0

最后我自己找到了答案。表名区分大小写，所以我根据我的记录更改了名称。问题解决了...

于 2017-12-12T20:04:07.493 回答