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PHP表單處理圖像和數據（偏差日誌）

2017-04-19 67 views
1

所以我被要求爲公司製作某種偏差日誌，這意味着每件錯誤的事情都需要注意。PHP表單處理圖像和數據（偏差日誌）

它需要一個唯一的ID，日期，文本，重要的是偏差的圖像。 我的問題是如果我如何以及如何製作表格並將其全部放入數據庫中，然後製作一個表格並使用上傳的圖片顯示數據。

<?php 
    if(isset($_POST['update'])) { 
     $dbhost = 'localhost'; 
     $dbuser = 'root'; 
     $dbpass = ''; 
     $tbl_name="heijsDB"; 

     $conn = mysql_connect($dbhost, $dbuser, $dbpass); 

     $volgnr = $_POST['volgnr']; 
     $constatering = $_POST['constatering']; 
     $radio = $_POST['radio']; 
     $actie = $_POST['actie']; 
     $date = $_POST['geplande_datum'] 
     $verant = $_POST['verantwoordelijke']; 
     $toelichting = $_POST['toelichting'] 



     if(! $conn) { 
      die('Could not connect: ' . mysql_error()); 
     }   
     $sql = "INSERT INTO logboek_afwijkingen (volgnr, ?IMAGE?, constatering, besmetting, actie, datum_gepland, verantwoordelijke, toelichting) VALUES ($volgnr, $constatering, ... , $radio, $actie, $date, $verant, $toelichting)" ; 
     mysql_select_db('HeijsDB'); 
     $retval = mysql_query($sql, $conn); 

     if(! $retval) { 
      die('Could not update data: ' . mysql_error()); 
     } 
     echo "Form send."; 

     mysql_close($conn); 
    }else { 
     ?> 
      <form method = "post" action = "<?php $_PHP_SELF ?>" id="form1"> 
    <fieldset id="f1"><b>Logboek afwijkingen</b></fieldset> 
     <table width= "600" border="1" cellspacing="1" cellpadding="1" class="w3-main w3-table"> 
      <tr><td width="200"><b>volgnummer</b></td> <td></td><td><input type="number" size="3" name="volgnr"></td></tr> 

      <tr><td width="200"><b>upload foto</b></td><td></td> <td width="100"><input type="file" accept="image/jpeg" size="3" name="picture"></td></tr> 


      <tr><td width="200"><b>constatering</b></td><td></td> <td width="100"><input type="" size="3"></td></tr> 

      <tr><td width="100"><b>kans op acute besmetting</b></td><td></td> <td><input type="radio" value="ja" class="w3-radio" name="radio"> Ja<br> 
                       <input type="radio" value="nee" class="w3-radio" name="radio"> Nee<br></td></tr> 
      <tr><td width="100"><b>corrigerende actie</b></td><td></td> <td><input type="textarea" size="3" name="actie"></td></tr> 
      <tr><td width="100"><b>geplande datum</b></td><td></td> <td><input type="date" size="3" name="geplande_datum"></td></tr> 

      <tr><td width="100"><b>verantwoordelijke</b></td><td></td> <td><input type="text" size="3" name="verantwoordelijke"></td></tr> 

      <tr><td width="100"><b>Toelichting</b></td><td></td> <td><input type="text" size="3" name="toelichting"></td></tr> 

     </table> 
      <input type="submit" name="update" value="update" action="<?php echo $_SERVER['PHP_SELF']; ?>"> 
     <?php 
    }

這是我的代碼，希望你們能幫幫我！

來源

2017-04-19 maus

回答

-1

您需要使用Handling file uploads。 在我看來，將圖像存儲在數據庫中並不是一個好主意。最好將上傳的文件放到目錄和數據庫中以存儲圖像路徑或文件名。

來源

2017-04-19 12:28:09 Psyhos

0

首先，添加圖像路徑進入database.when你想從數據庫中獲取圖像，然後使用查詢象下面這樣：

<?php 
    $retrieve = "SELECT image FROM tbl_name"; 
    $exe = mysqli_query($conn,$retrieve); 
?> 
<form method = "post" action = "<?php $_PHP_SELF ?>" id="form1" enctype="multipart/form-data"> 
    <fieldset id="f1"><b>Logboek afwijkingen</b></fieldset> 
     <table width= "600" border="1" cellspacing="1" cellpadding="1" class="w3-main w3-table"> 
      <tr><td width="200"><b>volgnummer</b></td> <td></td><td><input type="number" size="3" name="volgnr"></td></tr> 

      <tr><td width="200"><b>upload foto</b></td><td></td> <td width="100"><input type="file" accept="image/jpeg" size="3" name="picture"></td></tr> 


      <tr><td width="200"><b>constatering</b></td><td></td> <td width="100"><input type="" size="3"></td></tr> 

      <tr><td width="100"><b>kans op acute besmetting</b></td><td></td> <td><input type="radio" value="ja" class="w3-radio" name="radio"> Ja<br> 
                       <input type="radio" value="nee" class="w3-radio" name="radio"> Nee<br></td></tr> 
      <tr><td width="100"><b>corrigerende actie</b></td><td></td> <td><input type="textarea" size="3" name="actie"></td></tr> 
      <tr><td width="100"><b>geplande datum</b></td><td></td> <td><input type="date" size="3" name="geplande_datum"></td></tr> 

      <tr><td width="100"><b>verantwoordelijke</b></td><td></td> <td><input type="text" size="3" name="verantwoordelijke"></td></tr> 

      <tr><td width="100"><b>Toelichting</b></td><td></td> <td><input type="text" size="3" name="toelichting"></td></tr> 

     </table> 
</form>

來源

2017-04-19 12:11:51

0

得到它的工作，感謝您的意見！

<?php 
    if(isset($_POST['update'])) { 
     $dbhost = 'localhost'; 
     $dbuser = 'root'; 
     $dbpass = ''; 
     $tbl_name="heijsDB"; 

     $conn = mysql_connect($dbhost, $dbuser, $dbpass); 


     $constatering = $_POST['constatering']; 
     $radio = $_POST['radio']; 
     $actie = $_POST['actie']; 
     $date = $_POST['geplande_datum']; 
     $verant = $_POST['verantwoordelijke']; 
     $toelichting = $_POST['toelichting']; 
     $filename=$_FILES['file']['name']; 
     $filetype=$_FILES['file']['type']; 

    if($filetype=='image/jpeg' or $filetype=='image/png' or $filetype=='image/gif') 
{ 
     move_uploaded_file($_FILES['file']['tmp_name'],'uploads/'.$filename); 
     $filepath="uploads/".$filename; 

     if(! $conn) { 
      die('Could not connect: ' . mysql_error()); 
     }   
     $sql = "INSERT INTO logboek_afwijkingen (image, constatering, besmetting, actie, datum_gepland, verantwoordelijke, toelichting) VALUES ('$filepath', '$constatering', '$radio', '$actie', '$date', '$verant', '$toelichting')" ; 
     mysql_select_db('HeijsDB'); 
     $retval = mysql_query($sql, $conn); 

     if(! $retval) { 
      die('Could not update data: ' . mysql_error()); 
     } 
     echo "Form send."; 


     mysql_close($conn); 
    }}else { 
     ?> 
      <form method = "post" action = "<?php $_PHP_SELF ?>" id="form1" enctype="multipart/form-data"> 
    <fieldset id="f1"><b>Logboek afwijkingen</b></fieldset> 
     <table width= "600" border="1" cellspacing="1" cellpadding="1" class="w3-container w3-table"> 

      <tr><td width="200"><b>upload foto</b></td><td></td> <td width="100"><input type="file" size="3" name="file"></td></tr> 


      <tr><td width="200"><b>constatering</b></td><td></td> <td width="100"><input type="text" name="constatering" size="3"></td></tr> 

      <tr><td width="100"><b>kans op acute besmetting</b></td><td></td> <td><input type="radio" value="ja" class="w3-radio" name="radio"> Ja<br> 
                       <input type="radio" value="nee" class="w3-radio" name="radio"> Nee<br></td></tr> 
      <tr><td width="100"><b>corrigerende actie</b></td><td></td> <td><input type="textarea" size="3" name="actie"></td></tr> 
      <tr><td width="100"><b>geplande datum</b></td><td></td> <td><input type="date" size="3" name="geplande_datum"></td></tr> 

      <tr><td width="100"><b>verantwoordelijke</b></td><td></td> <td><input type="text" size="3" name="verantwoordelijke"></td></tr> 

      <tr><td width="100"><b>Toelichting</b></td><td></td> <td><input type="text" size="3" name="toelichting"></td></tr> 

     </table> 
       <input type="submit" name="update" value="update" action="<?php echo $_SERVER['PHP_SELF']; ?>"> 
     </form> 

     <?php 
    }

來源

2017-04-20 09:02:51 maus

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