Question

我想知道如何检查特定帖子的数据库是否没有输入标签，如果没有，则显示no tags have been entered。我怎么能用PHP做到这一点？

这是我的PHP＆amp; MySQL代码。

$tag = array();
$dbc = mysqli_query($mysqli, "SELECT tags.tag
                  FROM posts_tags 
                  INNER JOIN tags ON tags.id = posts_tags.tag_id 
                  WHERE posts_tags.post_id = '" . $post_id . "' 
                  GROUP BY tags.tag");

if (!$dbc) {
    print mysqli_error($mysqli);
}  else {
    while($row = mysqli_fetch_array($dbc)){
    $tag[] = $row['tag'];
    }
}

Answer 1

如果标签数组为空，则没有标签：

if(empty($tag)){
   echo 'no tags have been entered';
}

您还可以检查返回的行数：

if(mysqli_num_rows($dbc) == 0){
   echo 'no tags have been entered.';
}

Answer 2

使用mysqli_num_rows和代码如下：

$tag = array();
$dbc = mysqli_query($mysqli, "SELECT tags.tag
                  FROM posts_tags 
                  INNER JOIN tags ON tags.id = posts_tags.tag_id 
                  WHERE posts_tags.post_id = '" . $post_id . "' 
                  GROUP BY tags.tag");

if (!$dbc)
{
    print mysqli_error($mysqli);
}
elseif(!mysqli_num_rows($dbc))
{
    echo "no tags have been entered";
}
else
{
    while($row = mysqli_fetch_array($dbc))
    {
        $tag[] = $row['tag'];
    }
    print_r($tag);
}

Answer 3

$（＆＃34; .signup_name_image＆＃34）。隐藏（）;

var error_signup_name = false;

$(".signup_name").blur(function(){
    check_signup_name();
});

function check_signup_name() {
    var signup_name = $.trim($('.signup_name').val());
    if (signup_name == '' || signup_name.length < 6 || signup_name.length > 50 ) {
        $(".signup_name").css({"border":"1px solid #ff0000"});
        $(".signup_name_image").show();
        $(".signup_name_show").html("&#8226; Write Full Name in 6 to 50 alphabets");
        **$(".signup_name").focus();**
        error_signup_name = true;
    } else if (/^[a-zA-Z0-9- ]*$/.test(signup_name) == false){
        $(".signup_name").css({"border":"1px solid #ff0000"});
        $(".signup_name_image").show();
        $(".signup_name_show").html("&#8226; Contain invalid character");
        **$(".signup_name").focus();**
        error_signup_name = true;
    }else {
        $(".signup_name").css({"border":"1px solid #21cc00"});
        $(".signup_name_image").hide(); 
    }
}

$("#signup_form").submit(function(){

    error_signup_name = false;

    check_signup_name();

    if (error_signup_name == false) {
        return true;
    } else {
        return false;
    }
});

$(".signup_name_image").hover(function(){
    $(".signup_name_show").show();
}, function() {
    $(".signup_name_show").hide();}
);

现在福克斯被困在同一领域提交时如何关注字段

PHP＆amp; MySQL检查数据库是否输入数据问题

3 个答案: