php mysql下拉问题

Question

我正在尝试从我的sql数据库构建一个下拉列表.管理以使其工作有点,但列表只显示表中的最后一项,并在页面重新加载时我简要地将其打印在页面上:

[
 {"id":"1","zavod_ime":"PEIT"},
 {"id":"2","zavod_ime":"PTJM"},
 {"id":"3","zavod_ime":"PM"},
 {"id":"4","zavod_ime":"PN"},
 {"id":"5","zavod_ime":"BS"}
]

这是我单独的php文件,它处理数组中的连接和放置:

    $sqltran = mysqli_query($con, "SELECT id, zavod_ime FROM zavod")or die(mysqli_error($con));
    $arrVal = array();

    while ($rowList = mysqli_fetch_array($sqltran)) {

                    $namez = array(
                        'id'=> $rowList['id'],
                        'zavod_ime'=> $rowList['zavod_ime']
                        );      


                        array_push($arrVal, $namez);    
    }
         echo  json_encode($arrVal);        


    mysqli_close($con);

和html中的列表部分:

                    <select>
                        <?php foreach($namez as $zavod): ?>
                        <option value="<?= $namez['id']; ?>"><?= $namez['zavod_ime']; ?></option>
                        <?php endforeach; ?>
                    </select>  

想法？

Answer 1

你需要arrVal循环而不是namez

<select>
    <?php foreach($arrVal as $namez): ?>
    <option value="<?= $namez['id']; ?>"><?= $namez['zavod_ime']; ?></option>
    <?php endforeach; ?>
</select>