小编Ric*_*oss的帖子

所以这是我的代码.我试图设置隐藏表单输入的值.

if($gender == "BOY" || $gender == "MALE" || $gender == "M" || $gender == "1") {
        console.log('Its a boy!');
        $(".girlsonly").hide();
        document.getElementsByName('dressjacket')[0].placeholder='Jacket';
        $('#majorgender').val('1');
    };
    if($gender == "GIRL" || $gender == "FEMALE" || $gender == "F" || $gender == "0") {
        console.log('Its a girl!');
        document.getElementsByName('dressjacket')[0].placeholder='Dress';
        $('#majorgender').val('0');
    };

这是表单输入:

<input type="hidden" name="majorgender" id="majorgender"></input>

它成功地记录了"它是一个女孩!" 或者"它是个男孩!" 但之后当我插入" alert($('majorgender').val());"时它会提醒" 未定义 "

如果我记录alert($('majorgender'))它回复"对象对象",所以我知道它存在,我只是不知道为什么它没有正确设置.

有什么想法吗？

所以我正在使用AJAX上传一个图像,我正在构建这个图像管理器工具...突然间它无法工作......我没有更改任何代码或任何东西.

.php运行并上传图像,但是在json被编码并发送回来之后我想要触发的事件没有发生.:/

控制台记录 Uncaught SyntaxError: Unexpected token <

另一个AJAX请求也发生了同样的事情,但第三个问题就好了......

在过去,当我收到此错误时,这是​​一个PHP语法错误,但我还没有更新.php一段时间.. soooo我难倒.

这是我的Javascript:

$("#chosenfile").change(function(){

    if($(this).attr('name')) {

        var data;

        data = new FormData();
        data.append('uploadedfile', $( '#chosenfile' )[0].files[0]);
        $('#loader').css('display','block');
        $.ajax({
            url: '/includes/uploadimage.php',
            data: data,
            processData: false,
            contentType: false,
            type: 'POST',
            success: function ( data ) {
                //console.log(data); //returns a string of the data.
                var image = JSON.parse(data); //parses the string into an object.
                console.log(image); //logs the object.
                if (image.error) {
                    alert(image.error);
                    $('#remove').click();
                    $('#loader').css('display','none');
                } else {
                    if (image.link) { //If the image is …

这就是我得到的,更新的

$avg_query = "SELECT AVG(rating) AS avg_rating 
                FROM kicks 
               WHERE userid = '$userid'";

$avg_result = mysqli_query($cxn, $avg_query) or die("Couldn't execute query.");                         
$row = mysqli_fetch_assoc($avg_result);
$average = $row['avg_rating'];

每当我回复它时,它总是返回数组.从来没有使用过这个功能,但我不明白为什么会这样做.

我的代码是.

$newModel = "INSERT INTO models (id," . 
    " firstname," .
    " lastname," .
    " email," .
    " password," .
    " group," .
    " phone," .
    " timeofday," .
    " dayofweek," .
    " address," .
    " city," .
    " state," .
    " zip," .
    " gender," .
    " hair," .
    " eye," .
    " birthday," .
    " birthmonth," .
    " birthyear," .
    " bustshirt," .
    " cup," .
    " waist," .
    " hips," .
    " waist," .
    " hips," . …

这是我的代码:

$query = "SELECT * directory WHERE id ='$comId'";

echo $query;

$com_result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_assoc($com_result)) {  // While I am finding agents.

以下是$ query等于:

SELECT * directory WHERE id ='4'

这是我后来得到的:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'directory WHERE id ='4'' at line 1