Question

我有这个PHP代码从html表单（名称，文件，照片，地址，.....）中获取值，并尝试在sql数据库中插入或更新它们。

<?php 
$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name=""; // Table name 
session_start();
$con=mysqli_connect("localhost","","","");

$id=$_REQUEST['id']; 

//Variable intilisation 
$name = '';
$remarcs = '';
$address = '';
$test_res = '';
$date = '';
$phone = '';
$new_path = '';



if (isset ($_POST['name'])) {
    $name = $_POST['name'];
}
if (isset ($_POST['remarcs'])) {
    $remarcs = $_POST['remarcs'];
}
if (isset ($_POST['test_res'])) {
    $test_res = $_POST['test_res'];
}
if (isset ($_POST['address'])) {
    $address = $_POST['address'];
} 

if (isset ($_POST['date'])) {
    $date = $_POST['date'];
} 

if (isset ($_POST['phone_num'])) {
    $phone = $_POST['phone_num'];
}

if(!empty($_FILES)){ //Check file is uploaded or not 
    $check = getimagesize($_FILES["file"]["tmp_name"]);
    if($check !== false) {
        $path = "../uploads/".$_FILES['file']['tmp_name'];
        move_uploaded_file($file, $path);
        $new_path = $path;
        echo "Uploaded";
    } 
    if($check ==  false){
        echo "Not uploaded";
    }
}
if(move_uploaded_file($_FILES['file']['tmp_name'], $path)){
$sql="  update patients set values
            name = '$name',
            echo_photo = 'NULL',
            echo_file = '$new_path',
            remarcs = '$remarcs',
            test_res = '$test_res',
            date = '$date',
            address = '$address',
            phone_num = '$phone' 
        WHERE id = ".$id;

$result=mysqli_query($con,$sql) or die('Unable to execute query. '. mysqli_error($con));
header("location:update_done.php");

}else{
        header("location:update_false.php");

}
if($result){
    echo $name."<p>\n</p>";
    echo $remarcs."<p>\n</p>";
    echo $test_res."<p>\n</p>";
    echo $address."<p>\n</p>";
    echo $phone."<p>\n</p>";  
}

mysqli_close($con);

＆GT？;

问题是当我上传下图所示的文件时，我收到以下错误：

当我没有上传文件时，我收到了一个错误：请选择一个文件。有帮助吗？谢谢。

这是html表单：

<form action="update.php" id="Form2" method="POST" enctype="multipart/form-data" class="c">
<div align="center">
<?php echo "Updating information about Patient ".$row["name"]; ?>
<table class="imagetable" border="1" cellspacing="3"  align="center">
  <th>Personal Informations</th>
<th>Test Results</th>

<tr><td>Name<br>
  <input type="text" class="large-fld" name="name" placeholder="Patient Name" value="<?php echo $row['name'];?>"/></td>
<td>Remarcs:<br>
  <textarea type="text" cols="40" rows="5" class="large-fld"  name="remarcs" placeholder="Remarcs"><?php echo $row['remarcs'];?></textarea></td>
  <tr><td>Address<br>
  <input type="text" class="large-fld" name="address" placeholder="Address" value="<?php echo $row['address'];?>"/>
  </td>
<td>Test<br>  <textarea type="text" cols="40" rows="5" class="large-fld"  name="test_res" placeholder="Test Result"><?php echo $row['test_res'];?></textarea></td></tr>
</td>
</tr>
<tr><td>Phone Number<br>
  <input type="text" class="large-fld" name="phone_num" placeholder="Phone Number" value="<?php echo $row['phone_num'];?>"/>
  </td>
<th>Files</th>
</tr>
  <td>Scanned Echo Photo<br>
  <input type="file" class="" name="echo_photo" id="echo_photo" placeholder="Add echo photo" value="<?php echo $row['echo_photo'];?>"/></td>
<td>Echo Files:<br>
      <input type="file" name="file" id="file"  value="<?php echo $row['echo_files'];?>"/><br></td>
  </tr></th></table>
  <div class="row" align="center">
<input type="submit" name="submit" id="btnUploadId" class="large-btn" value="Update" onClick="btnOnClickUpload()">
      <input type="hidden" id="courseIdHidden" value="<?php echo $idd; ?>" /></td></tr>
      </table></div>
</form>

Answer 1

尝试初始化变量。我添加了整个代码。你不需要获得$ _POST [＆＃39; file＆＃39;]。

<?php 
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="clinic"; // Database name
$tbl_name="patients"; // Table name
session_start();
$con=mysqli_connect("localhost","root","root","clinic");

$id=$_REQUEST['id'];

 $name = '';
 $remarcs = '';
 $address = '';
 $test_res = '';
 $date = '';
 $phone = '';
  $new_path = '';

if (isset ($_POST['name'])) {
 $name = $_POST['name'];
 }
if (isset ($_POST['remarcs'])) {
 $remarcs = $_POST['remarcs'];
 }
if (isset ($_POST['test_res'])) {
$test_res = $_POST['test_res'];
}
if (isset ($_POST['address'])) {
  $address = $_POST['address'];
}

if (isset ($_POST['date'])) {
  $date = $_POST['date'];
}

if (isset ($_POST['phone_num'])) {
 $phone = $_POST['phone_num'];
}

if(isset($_FILES['file'])){ //Check file is uploaded or not

  $path = "../uploads/".$_FILES['file']['name'];
  if(move_uploaded_file($_FILES["file"]["tmp_name"], $path)){
  $new_path = $path;
  $sql=" update patients set 
  name = '$name',
  echo_photo = 'NULL',
  echo_file = '$new_path',
  remarcs = '$remarcs',
   test_res = '$test_res',
  date = '$date',
  address = '$address',
  phone_num = '$phone'
  WHERE id = ".$id;

  $result=mysqli_query($con,$sql) or die('Unable to execute query. '.   mysqli_error($con));

if($result){
 echo $name."<p>\n</p>";
 echo $remarcs."<p>\n</p>";
 echo $test_res."<p>\n</p>";
 echo $address."<p>\n</p>";
 echo $phone."<p>\n</p>";
 }
  echo "Uploaded";
} else {
  echo "Not uploaded";
 }
}
mysqli_close($con);

?>

Answer 2

这是最终的工作代码：

if (isset ($_POST['name'])) { 
$name = $_POST['name']; 
} 
if (isset ($_POST['remarcs'])) { 
$remarcs = $_POST['remarcs']; 
} 
if (isset ($_POST['test_res'])) { 
$test_res = $_POST['test_res']; 
} 
if (isset ($_POST['address'])) { 
$address = $_POST['address']; 
} 

if (isset ($_POST['date'])) { 
$date = $_POST['date']; 
} 

if (isset ($_POST['phone_num'])) { 
$phone = $_POST['phone_num']; 
} 

if(isset($_FILES['file'])){ //Check file is uploaded or not 

$path = "../uploads/".$_FILES['file']['name'];
//$path2 = "../uploads/".$_FILES['echo_photo']['name']; 

if(move_uploaded_file($_FILES["file"]["tmp_name"], $path)){ 
$new_path = $path; 
$sql="UPDATE $tbl_name SET  
name = '$name', 
echo_files = '$new_path', 
remarcs = '$remarcs',
test_res = '$test_res', 
date = '$date', 
address = '$address', 
phone_num = '$phone' 
WHERE id = '$id'"; 

$result=mysqli_query($con,$sql) or die('Unable to execute query. '. mysqli_error($con));

将blob文件/图像上传到Mysql

2 个答案: