Question

嘿伙计们我已经创建了一个带有图像上传的注册表单，但是当我尝试更新此表单时，我尝试获取id但是我的图像的isset无效，所以它只是不会运行我的更新查询请检查它出

这是更新表单，其中所有值都将显示以供编辑现在我可以在提交按钮的isset条件下运行更新功能，然后更新数据

    <title>Register Update</title>
<?php
//error_reporting(0);
$id=$_GET['id'];

function __autoload($classname)
{
include "$classname.php"; 
}
$obj = new connect();
$st=$obj->con();

if (isset($_POST['sub'])) 
{
$upd= new update();
$upd->updatedata($_POST);
}





$qry = "select * from register ";
$run = mysqli_query($st,$qry);
$row = mysqli_fetch_assoc($run);
 {
 $g = $row['gen'];
 $l = $row['lang'];
}


$query=mysqli_query($st,"select * from register where id='$id'");
//echo "<ul>";
while($query2=mysqli_fetch_assoc($query))
{
    //print_r($query2);
echo "<form method='POST' action='RegisterRetrieve.php'>";
echo "<table>";
?>

<p><input type="hidden" name="sid" value="<?php echo $query2['id']; ?>"></p>
<tr>
<td>
First Name:
</td>
 <td><input type="text" name="uname" value="<?php echo $query2['uname']; ?>"></td>
</tr>
<tr>
<td>Password:</td>
    <td><input type="password" name="pwd" value="<?php echo $query2['pwd']; ?>"></td>
</tr>
<tr>
<td>Email Id:</td>    
<td><input type="text" name="emailid" value="<?php echo $query2['emailid']; ?>"
</td>
</tr>
<tr>
<td>Radio Button: Are you male or female?</td>
    <?php
    if ($g == "male"){
echo "<td><input type='radio' name='gen' value='Male' id='gen' checked> Male <input type='radio' name='gen' value='Female' id='gen'> Female </td>";
 }
 else
 {
 echo "<td><input type='radio' name='gen' value='Male' id='gen'> Male <input type='radio' name='gen' value='Female' id='gen' checked> Female </td>";
 }
    ?>
</tr>
<tr>    
<td>Check Box: Check the languages you know?</td>
<td><?php
    $lang=explode(',',$l); 
    //print_r($lang);
        if(in_array('Cricket', $lang))
            echo '<input type="checkbox" name="lang[0]" value="Cricket" checked>Cricket';
        else
            echo '<input type="checkbox" name="lang[0]" value="Cricket">Cricket';
        if(in_array('Basketball', $lang))
            echo '<input type="checkbox" name="lang[1]" value="Basketball" checked>Basketball';
        else
            echo '<input type="checkbox" name="lang[1]" value="Basketball">Basketball';

        if(in_array('Hockey', $lang))
            echo '<input type="checkbox" name="lang[2]" value="Hockey" checked>Hockey';
        else
            echo '<input type="checkbox" name="lang[2]" value="Hockey">Hockey'."<br>";  
    ?>   
</td>
</tr>
<tr>
<td>Mobile No:</td>
<td><input type="text" name="mobile" value="<?php echo $query2['mobile']; ?>"
</td>
</tr>
<tr>
<td>10th Marks:</td>
<td><input type="text" name="marks_10" value="<?php echo $query2['10marks'];?>"
</td>
</tr>
<tr>
<td>
12th Marks:</td>
<td><input type="text" name="marks_12" value="<?php echo $query2['12marks'];?>"</td>
</tr>
<tr>
<td>
Browse Image:</td>
<td><input type="file" name="file1"></td>
<td><img src='img/<?php echo $query2['name'];?>' width='150px' height='150px'></td>
</tr>
<tr>
<td>
<select name="priority">
<option value="admin">
admin
</option>
<option value="<?php echo $query2['priority']; ?>"><?php echo $query2['priority']; ?>
</option>
<option value="superadmin">
superadmin
</option>

</select>
</td>
</tr>
<tr>
<td></td>
<td>
<input type="submit" value="submit" name="sub"><br>
</td>
</tr>
<?php

echo "<table>";
echo "</form>";
}
//echo "</ul>";



?>

现在我的更新查询我正在使用但当我尝试设置我的图像时它就不会进入那种情况

    <?php

class update extends connect
{

    function updatedata($rel)
    {
        $obj= new connect();    
        $obj->con();


        extract($_POST);

        $id=$_GET['id'];

        $line = implode("," ,$lang);
        print_r($_POST);   


        if(isset($_FILES["file1"]))
        {

        extract($_POST);
        echo "hello";
        $name = $_FILES['file1']['name'];

        $type = $_FILES['file1']['type'];

        $size = $_FILES['file1']['size'];

        $tmp_name = $_FILES['file1']['tmp_name'];

        $loc = 'img/';

        $ext = substr($name,strpos($name,'.')+1);

        if($_FILES['file1']['size']>= '10000' || $_FILES['file1']['size']<="23000000")
        {
            //echo $size;
        }
            else{
            //  echo "size is not supported";
            }
        $val = $_FILES['file1']['size'];

         if($ext == 'jpg' || $ext == 'png')
            {
    //echo $lang;
    //print_r($_POST);
    //exit;
    $val =("update register set uname='$uname',pwd='$pwd',emailid='$emailid',gen='$gen',lang='$line',mobile='$mobile',10marks='$marks_10',12marks='$marks_12' file1='$name' where id=$sid");
    //print_r($qry);

    $res=mysqli_query($this->con(),$val);
    //print_r($run);
    if($res)
    {
            move_uploaded_file($tmp_name,$loc.$name);
            //echo "data saved";
        //echo "Data inserted";

        }
        else
        {
            //echo "Data Not Inserted";

            }

    }
}
}


        //print_r($val);

    //  return $res;
    }
?>

Answer 1

你的HTML坏了：

</tr>
<?php

echo "<table>";    <--shouldn't this be </table>?
echo "</form>";
}

我试图用图像更新来更新我的数据，但图像设置将无法在PHP中工作

1 个答案: