Question

我在PHP代码中收到“Notice：Undefined index：k”。 K是文本字段的名称，我使用$ _GET []方法来获取k的值。在下面提到的示例中，我试图在提交表单后保持值可用。此代码第一次正常运行，但第二次出现上述错误。

    <form name="keywordquery" method="get" action="keywordsearch.php">
<fieldset class="fieldsetclass"><legend class="legendclass">Search by 
     Keywords</legend>
    <div id="searchbox">
   <input type="text" name="k" value="<?php if(isset($_GET['k'])){echo  
   htmlentities($_GET['k']);} ?>" style="border: 1px, thin; width:92%; "/> <input     
    type="image" style="margin-bottom: 0; margin-top: 2px;" src="search.png" 
    value="submit" />
</div>
</fieldset>
    </form>
   </div>
  <table id="dataTable" cellpadding="0" cellspacing="0" border="1" style="border-    
   radius:20px; box-shadow: 9px 5px 8px #7E9044;">
  <tbody>    
<?php
        // PAGINATION Code. check if the starting row variable was passed in the 
    URL or not
  if (!isset($_GET['startrow']) or !is_numeric($_GET['startrow'])) {

  //we give the value of the starting row to 0 because nothing was found in URL
  $startrow = 0;
  //otherwise we take the value from the URL
   } else {
  $startrow = (int)$_GET['startrow'];
   }


$k1 = $_GET['k'];

$term = explode(" ", $k1);
$query = "SELECT * FROM data ";
foreach ($term as $each) {
$i++;
    if($i==1)
    {
        $query .= "WHERE keywords LIKE '%$each%' ";
    }

else {
    $query .= " OR WHERE keywords LIKE '%$each%' ";
}

}

$query .= "LIMIT $startrow, 1";

$connection = mysql_connect("xxxx", "xxxxx","");
if(!$connection)
echo "No database connected";
$dbase = mysql_select_db("xxxxxxxx", $connection);
if(!$dbase)
echo "No datatable connected";
$ourquery1 = mysql_query ($query);
if(!$ourquery1)
echo "No query found";

$row1 = mysql_num_rows ($ourquery1);
if ($row1 > 0)
{

    while($result = mysql_fetch_assoc($ourquery1))
    {
        echo "<tr>";
        $title = $result['title'];
        $link = $result['link'];
        $region = $result['region'];
        $sector = $result['sector'];
        $theme = $result['theme'];      
        echo "<td> <a href=$link><h3>$title<h3></a>";
        echo "<h4>Sector: $sector&nbsp; &nbsp; &nbsp; 
                    &nbsp; &nbsp; &nbsp; Theme: $theme &nbsp; 
        &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<br> Region: $region </td>";
    }
}


    echo "</tbody>";
       echo '<a href="'.$_SERVER['PHP_SELF'].'?startrow='.($startrow+1).'">Next</a>';
       echo '<a href="'.$_SERVER['PHP_SELF'].'?startrow='.($startrow-1).'">Prev</a>';

Answer 1

替换该行： $ k1 = $ _GET ['k'];

有类似的东西： $ k1 = isset（$ _ GET ['k']）？ $ _GET ['k']：$ default_k_value;

Answer 2

你没有显示完整的表格，所以很难说出错是什么，但这里有一个提示。使用$ _REQUEST交换$ _GET。

示例：

<?php if(isset($_REQUEST['k'])){echo  
   htmlentities($_REQUEST['k']);} ?>

如果表单使用POST方法，则值将在$ _POST中。如果表单使用GET方法，则值将在$ _GET中。但是，$ _REQUEST将包含表单字段，无论表单是使用POST还是GET。

变量的PHP值未定义

2 个答案: