小编Cod*_*die的帖子

我试图使用PHP表单来更新SQL数据库.

花了好几个小时后我终于能够更新数据库,但即使找不到与字符串匹配的行,问题仍然存在,它仍然会返回消息"已成功更新数据".我检查了数据库,并没有输入任何其他的输入,但我真的想让用户知道他们错误输入ID.

谢谢

if (isset($_POST['update'])) {
    $dbhost = 'localhost';
    $dbuser = 'user';
    $dbpass = 'password';
    $conn = mysql_connect($dbhost, $dbuser, $dbpass);

    if (! $conn ) {
        die('Could not connect: ' . mysql_error());
    }

    $UniqueID = mysql_real_escape_string($_POST['UniqueID']);
    $Referral = mysql_real_escape_string($_POST['Referral']);

    mysql_select_db('database');

    $sql = "UPDATE tablename set Referral = '$Referral' WHERE ID = '$UniqueID'"; 

    $retval = mysql_query( $sql, $conn );

    if(! $retval ) {
        die('Could not update data: ' . mysql_error());
    } else {
        echo "Updated data successfully\n";
    }

   mysql_close($conn);
}

谢谢

我正在尝试从我的数据库中获取数据并将它们存储在多维数组中,但由于某种原因它没有返回任何行.在查看我的代码之前,我想解释一下我希望如何将数据存储在多维数组中.所以,例如,这是一个表:

Name  | Age | Sex
Harry | 18  | Male
Kate  | 18  | Female

从数据库中获取这些后,我想将它们存储在多维数组中,如下所示:

$array[0][0] = "Harry";
$array[0][1] = "18";
$array[0][2] = "Male";
$array[1][0] = "Jane";
$array[1][1] = "18";
$array[1][2] = "Female";

代码部分

<?php

include("connect.php");

global $connect;
$fetch = mysqli_query($connect,"SELECT * FROM allData limit 2");
$count = mysqli_num_rows($fetch);

while($row=mysqli_fetch_array($fetch,MYSQLI_NUM)){

  $count --;
  $arrayCount = count($myArray);

  $tempArrayCount = 0;
  while($tempArrayCount < $arrayCount){
  $array[$count][$tempArrayCount]= $row[$tempArrayCount];
  $tempArrayCount++;
  }

}

print_r($array);

?>

当我的时候,我的阵列看起来像这样 print_r(delegates[$i])

Array
(
    ['firstName'] => Brady
    ['surname'] => Manning
    ['jobTitle'] => CEO
    ['memberNumber'] => 123456
    ['dieteryRequirements'] => halal
)

然后,当我尝试获取firstName时,echo delegates[$i]['firstName']我得到以下错误

Notice: Undefined index: firstName 

小学生的问题,但我真的很感激这里的一些帮助.

我有一个PHP代码如下,但无法显示具有正确的PHP值的标题例如在独立的基础上它显示正确的数据为

$this->CatName; - Displays category

$this->prodDet->prod_name; - Displays Product Name

$this->prodDet->v_location; - Displays Location

我想创建一个组合标题

Used <catname> <prod_name> for sale in <v_location>

Like
Used Fisher Milk Bottle for sale in Capetown

但是当我修改代码时

  $title = 'Used' ($this->CatName $this->prodDet->prod_name) 'for Sale in' ($this->prodDet->v_location);

它显示语法错误,但不起作用

这是完整的代码

<?php
defined('_JEXEC') or die;

jimport('joomla.application.component.view');




/**
* Prepares the document
*
* @return void
*
* @throws Exception
*/
protected function _prepareDocument()
{
    $app = JFactory::getApplication();
    $menus = $app->getMenu();
    $title = null;

    $menu = $menus->getActive();

    if ($menu) …