Question

我最近决定使用较新版本的PHP测试我真正的旧项目，因此我必须从mysqli_fetch_assoc()切换到<?php $query = 'SELECT people_id, people_fullname FROM people WHERE people_isactor = 1 ORDER BY people_fullname;'; $result = mysqli_query($con,$query) or die(mysqli_error($con)); //populate the select options with the result while($row = mysqli_fetch_assoc($result)){ foreach($row as $value){ if($row['people_id'] == $movie_leadactor){ echo '<option value="'.$row['people_id'].'"selected="selected">'; }else{ echo '<option value="'.$row['people_id'].'">'; } echo $row['people_fullname'].'<option>'; } } ?>。

但是，在我的localhost上运行它时，数据将被返回两次。

以下是代码段。任何帮助将不胜感激。

提前致谢：）

Link

Answer 1

1.删除foreach()

2.Last <option>必须为</option>。

如下所示： -

<?php
$query = 'SELECT people_id, people_fullname FROM people WHERE people_isactor = 1 ORDER BY people_fullname;';
$result = mysqli_query($con,$query)  or die(mysqli_error($con));

//populate the select options with the result
while($row = mysqli_fetch_assoc($result)){

    if($row['people_id'] == $movie_leadactor){
        echo '<option value="'.$row['people_id'].'"selected="selected">'.$row['people_fullname'].'</option>';
    }else{
        echo '<option value="'.$row['people_id'].'">'.$row['people_fullname'].'</option>';
    }
}
?>

mysqli_fetch_assoc（）返回两次数据

1 个答案: