我可以通過基於文本框中的值與表中的列中的值之間的匹配關系的表單，將值添加到MySQL表中嗎？

Question

我有一個包含許多列的MySQL表，其中包括一個名為“名稱”的列。

我想為一個包含兩個文本框A和B的網頁設計一個表單。在文本框A中，要求用戶輸入其名稱，在文本框B中，需要輸入一些其他信息。

我希望PHP腳本檢查Textbox A中的Name是否與MySQL表的Name列中已有的值匹配，如果匹配，則將Textbox B中的值添加到該表的另一列中。 如果未找到名稱，我希望腳本返回錯誤，類似於“在我們的數據庫中未找到您的預訂”。

是否可以使用PHP / MySQL進行此操作，如果可以，我將如何處理？

當前代碼

$row_count = count($_POST['name']);

mysql_select_db($database, $connection);

if ($row_count > 0) {

$values = array();

 for($i = 0; $i < $row_count; $i++) {
     // variable sanitation...
     $name = mysql_real_escape_string(ucwords($_POST['name'][$i]));
     $workshop = mysql_real_escape_string($_POST['workshop'][$i]);

     $query = "SELECT * FROM conference WHERE Name = '$name' ";
     $result = mysql_query($query);

     if ($result) {
     $rowcount = mysql_num_rows($result);

     if ($rowcount == 0) {echo "no bookings found"; }
     else {
     $row = mysql_fetch_row($result);
     $sql = "UPDATE conference SET Workshop = '$workshop' WHERE Name = '$name'";
     mysql_query($sql);
     }

     }
}

}

Answer 1

這應該使您入門。

//Get the value from the textbox
$name = mysql_real_escape_string($_POST['name']);
//Select all rows with the name
$query = "SELECT other_info FROM table1 WHERE name = '$name' ";
$result = mysql_query($query);
if ($result) {
  $rowcount = mysql_num_rows($result);
  //No rows found
  if ($rowcount == 0) {echo "no bookings found"; }
  else {
    //process the data.
    $row = mysql_fetch_row($result);
    $other_info = $row['other_info'];
    echo "other_info = ".hmtlentities($other_info);

參見此處： http : //php.net/manual/zh/book.mysql.php

對代碼的注釋

$row_count = count($_POST['name']);
if ($row_count > 0) {

  mysql_select_db($database, $connection);
  //$values = array(); 
  $name = array();
  $workshop = array()
  $replace = array()

  for($i = 0; $i < $row_count; $i++) {
    // variable sanitation...
    $name[i] = mysql_real_escape_string(ucwords($_POST['name'][$i]));
    $workshop[i] = mysql_real_escape_string($_POST['workshop'][$i]);
    //if you use option 2
    $replace[i] = "('".$name[i]."','".$workshop[i]."')";
  }
  $names = "'".implode("','",$name)."'";
  $query = "SELECT 1 FROM conference WHERE Name IN $names "; 
  $result = mysql_query($query);

  if ($result) {
    $rowcount = mysql_num_rows($result);

    if ($rowcount == 0) {
      echo "<HTMLCODE HERE>"."no bookings found"."<MORE HTML>"; 
    //OPTION 1
    } else {
      for($i = 0; $i < $row_count; $i++) {
        //$row = mysql_fetch_row($result);  
        $sql = "UPDATE conference SET Workshop = '$workshop[i]' 
                WHERE Name LIKE '$name[i]'";
        mysql_query($sql);
      }
    }
    //OPTION2
    } else {
      $replacement = implode(",",$replace);
      $sql = "REPLACE INTO conference (names, workshop) VALUES $replacement "
      mysql_query($sql);
    }
  }
}