Question

我正在处理用户的自定义报告。在此页面中，我为数据库中的选择数据提供了一个用户复选框选项。用户可以选择他想要从表中查看的数据。我的表有很多字段，如名称，阻止，联系..等等。如果用户只想看到uid和名字，那么他只会检查姓名和电话的方框。我的复选框是

       <form action="#" method="post">
 <input type="checkbox" name="check_list[]" value="uid"><label>uid</label>
<input type="checkbox" name="check_list[]" value="name"><label>name</label>
<input type="checkbox" name="check_list[]" value="block"><label>block</label>....and so more
 <input type="submit" name="submit" value="Submit"/>
</form>

和我正在尝试的PHP代码是

    if(isset($_POST['submit'])){

    $sql[] = "select ";

    foreach($_POST['check_list'] as $selected){
        $sql[]=$selected.',';
    }  

    $sql[]= " from emitra_basic LIMIT 5";

    foreach ($sql as $user) {

        $ram="$user";
    }

    if ($result=mysqli_query($conn, $ram))
    { 
        while ($row=mysqli_fetch_row($result))
        { 
            echo "id".$row['uid']."<br/>";
            echo "name".$row['name']."<br/>";
        }
    }
}

但没有结果。我该怎么办？

Answer 1

请尝试

$sql[] = "select ".implode(', ', $_POST['check_list'])." from emitra_basic LIMIT 5";

而不是

$sql[] = "select ";
foreach($_POST['check_list'] as $selected){
$sql[]=$selected.',';}  
$sql[]= " from emitra_basic LIMIT 5";

得到类似的东西：

select uid, name, block from emitra_basic LIMIT 5;

否则在最后一个字段的末尾会有一个,：

select uid, name, block, from emitra_basic LIMIT 5; -- This is not working!

Answer 2

尝试这样： - 编码没有大的改变

if(isset($_POST['submit'])){

$sql=array();

$sql[] = "select ";

$sql[]=join(",", $_POST['check_list']);

$sql[]= " from emitra_basic LIMIT 5";

//print_r($sql);

foreach ($sql as $user) {

    //echo $user."<br>";

    $ram.=$user;
}

//echo $ram; exit;

//select uid, from emitra_basic LIMIT 5

if ($result=mysqli_query($conn, $ram))
{ 
    while ($row=mysqli_fetch_row($result))
    { 
        echo "id".$row['uid']."<br/>";
        echo "name".$row['name']."<br/>";
    }
}

}

Answer 3

在php中尝试这个

if(isset($_POST['submit'])){
$sql1 = "select ";
$sql2 = "";
$checklist = $_POST['check_list'];
$count = count($checklist);
if($count!=1){
    $count = $count-1;
    for ($i=0; $i < $count ; $i++) { 
        $sql2 = $sql2.$checklist[$i].",";
    }
    $sql2 = $sql2.$checklist[$count];
} else{
    $sql2 = $checklist[0];
}
$sql3= " from emitra_basic LIMIT 5";
$totalSql =  $sql1.$sql2.$sql3;

 if ($result=mysqli_query($conn,$totalSql))
     { while ($row=mysqli_fetch_row($result))
     { echo "id".$row['uid']."<br/>";
echo "name".$row['name']."<br/>";}}

}

使用php中的复选框从mysql中选择数据库中的数据

3 个答案: