2

我是那些背景一直支持而不是开发的相对 php / mysql 新手之一。然而，由于像这样的优秀资源网站，我已经设法将我最想实现的许多事情放在一起，因此提前感谢所有过去、现在和未来的人。但是，我要么无法构建合适的谷歌查询来查找我需要的信息，要么在我认为的某个地方遗漏了一个简单的技巧。

简而言之，我有一个带有大约 15 列（字段）的表的 mysql 数据库。我有一个 php 前端，我可以使用它来查询数据库，或者返回所有行，或者只返回那些包含一般搜索项目的行，跨越所有字段。一切都很好，显示很好，效果很好。但是，我想让用户选择只选择他们希望在结果中返回的那些列（字段）。因此，我有一个搜索页面，其中显示了列名，并相应地分配了复选框和值（列名）。代码如下：

<form action="result.php" method="post">     Search for* :  <input type="text" name="searchterm" />
<input type="submit" name="submit" value="Search Database" /> <br /><br /><br />
<input type=button value="Select ALL" onclick="select_all( check, true ); return false;">
<input type=button value="Clear ALL" onclick="select_all( check, false ); return false;">
<table>
<tr><th>FirstName<br/><input type="checkbox" id="check" name="field[]" value="FirstName,"></th> 
<th>LastName<br/><center><input type="checkbox" id="check" name="field[]" value="LastName,"></center></th> 
<th>Age<br/><center><input type="checkbox" id="check" name="field[]" value="Age,">        </center></th> 
<th>Instrument<br/><center><input type="checkbox" id="check" name="field[]" value="Instrument,"></center></th>
<th>Grade<br/><center><input type="checkbox" id="check" name="field[]" value="Grade,"></center></th> 
<th>LandlineNo<br/><center><input type="checkbox" id="check" name="field[]" value="LandlineNo,"></center></th>
<th>MobileNo<br/><center><input type="checkbox" id="check" name="field[]" value="MobielNo,"></center></th> 
<th>Email<br/><center><input type="checkbox" id="check" name="field[]" value="Email,"></center></th> 
<th>AddressLine1<br/><center><input type="checkbox" id="check" name="field[]" value="AddressLine1,"></center></th> 
<th>AddressLine2<br/><center><input type="checkbox" id="check" name="field[]" value="AddressLine2,"></center></th> 
<th>Town<br/><center><input type="checkbox" id="check" name="field[]" value="Town,"></center></th> 
<th>County<br/><center><input type="checkbox" id="check" name="field[]" value="County,"></center></th> 
<th>Postcode<br/><center><input type="checkbox" id="check" name="field[]" value="Postcode,"></center></th> 
<th>Price<br/><center><input type="checkbox" id="check" name="field[]" value="Price,"></center></th> 
<th>Paid<br/><center><input type="checkbox" id="check" name="field[]" value="Paid,"></center></th> 
<th>LastUpdated<br/><center><input type="checkbox" id="check" name="field[]" value="LastUpdated,"></center></th>
</tr>
</table>
</form>

所有复选框都被指定为 field[] 作为名称，因此输出到 result.php 中的数组。经过一点进一步阅读后，我可以让数组以逗号分隔显示（回显）选定的列，每个：

for ($i=0; $i<count($_POST['field']); $i++){ 
$columns = addslashes($_POST['field'][$i]); 
echo $columns;
}

// output example

LastName,Age,Instrument,Email

但是，我现在不知道如何只在结果视图中检索选定的列。任何人都可以帮助我并朝着正确的方向发展。我有这个，但这不起作用，因为所有列都返回了。

$result = mysql_query("select * from TABLE where FirstName like '%$searchterm%' or LastName like '%$searchterm%' or Age like '%$searchterm%' or Instrument like '%$searchterm%' or Grade like '%$searchterm%' or LandlineNo like '%$searchterm%' or MobileNo like '%$searchterm%' or Email like '%$searchterm%' or AddressLine1 like '%$searchterm%' or AddressLine2 like '%$searchterm%' or Town like '%$searchterm%' or County like '%$searchterm%' or Postcode like '%$searchterm%' or Price like '%$searchterm%' or Paid like '%$searchterm%'");

echo '<table>';
while ($row = mysql_fetch_assoc($result)) {
if (empty($columns)) {
    $columns = array_keys($row);
    echo '<tr><th>'.implode('</th><th>', $columns).'</th></tr>';
}
$resultset[] = $row;
echo '<tr><td>'.implode('</td><td>', $row).'</td></tr>';
}
echo '</table>';

最后一步也是根据结果集检索和填充表的列标题。提前感谢您的任何指导、建议和改进。

吉姆

4

1 回答 1

0 
$result = mysql_query("select * from TABLE where FirstName like '%$searchterm%' or LastName like '%$searchterm%' or Age like '%$searchterm%' or Instrument like '%$searchterm%' or Grade like '%$searchterm%' or LandlineNo like '%$searchterm%' or MobileNo like '%$searchterm%' or Email like '%$searchterm%' or AddressLine1 like '%$searchterm%' or AddressLine2 like '%$searchterm%' or Town like '%$searchterm%' or County like '%$searchterm%' or Postcode like '%$searchterm%' or Price like '%$searchterm%' or Paid like '%$searchterm%'");
$start = FALSE;
echo '<table><thead><tr>';
while ($row = mysql_fetch_assoc($result)) {
foreach ($row as $column => $value)
{
// Now we print the columns if they aren't printed
if ($start === FALSE && !empty($value))
{
echo "<th>$column</th>";
}
}
// And if the columns were printed now we turn start to TRUE.
if ($start === FALSE)
{
echo "</tr></thead><tbody>";
$start === TRUE;
}
echo "<tr>";
// Now for each key and value in the array we verify if the user wanted to print the column and we print it.
foreach ($row as $column => $value)
{
if (!empty($value))
{
echo "<td>$value</td>";
}
echo '</tr>';
}
}
echo '</tbody></table>';

这应该有效。你可以在评论中看到它是如何工作的。

于 2012-07-24T12:50:58.003 回答