用下拉菜单在php中搜索表单

Question

美好的一天！ 我只是php编程的新手，所以我需要有人来帮助我。 我仍然找不到此代码出了什么问题。 我创建了带有下拉菜单的搜索表单以过滤结果。 但是当我进行搜索时什么也没有发生。 它只是刷新页面或显示错误标准消息。 任何答案将不胜感激！ ：）提前致谢！ 这是代码：

   <html>
    <head>
<basefont face="Arial">
</head>
<body>

<?php
error_reporting(E_ALL); 
if (!isset($_POST['Submit'])) {
// form not submitted
?>

<form action="<?=$_SERVER['PHP_SELF']?>" method="post">
search  
  <input type="text" name="search">
   <select size="1" name="dropdown">
    <option value="" selected>search By...</option>
    <option value="first">Company</option>
    <option value="last">Address</option>  
  </select>
  <input type="Submit" value="Submit" name="Submit"> 
</form>

<?php
}

else {

// Server Variables
$host = "localhost";
$user = "mdti";
$pass = "tnet";
$db = "ojt";

$search = empty($_POST['search'])? die ("ERROR: Enter search Criteria") : mysql_escape_string($_POST['search']);
$dropdown = empty($_POST['dropdown'])? die ("ERROR: Select from dropdown") : mysql_escape_string($_POST['dropdown']);


// Open Connection

$connect = mysql_connect($host, $user, $pass) or die ("Unable to connect to host");

//Select Database

mysql_select_db($db) or die ("Unable to connect to database");

//Create query

$query = "SELECT arCompanyname, arAddress FROM ar WHERE $dropdown like'$search'" or die (mysql_error());


$result = mysql_query($query) or die (mysql_error());

$num=mysql_numrows($result);

mysql_close($connect);

echo "<b><center>Database Output</center></b><br><br>";

$i=0;
while ($i < $num) {

$company=mysql_result($result,$i,"arCompanyname");
$address=mysql_result($result,$i,"arAddress");


echo "<br>Company: $company<br><br>Address: $address<hr><br>";

$i++;

}
}
?>
</body>
</html>

Answer 1

查看代码，最初的想法是：

change like'$search' with like '%$search%'

change mysql_numrows to mysql_num_rows

check the dropdwon option values 'arCompanyname' and 'arAddress' matching your field names.

change <?=$_SERVER['PHP_SELF']?> to <?php echo $_SERVER['PHP_SELF']; ?>

Answer 2

您在不打开mysql_connect情况下调用mysql_real_escape_string 。您需要在mysql_real_escape_string代码上方调用连接代码

// Open Connection

$connect = mysql_connect($host, $user, $pass) or die ("Unable to connect to host");

$search = empty($_POST['search'])? die ("ERROR: Enter search Criteria") : mysql_real_escape_string($_POST['search']);
$dropdown = empty($_POST['dropdown'])? die ("ERROR: Select from dropdown") : mysql_real_escape_string($_POST['dropdown']);

另外，您需要更改选择选项的值

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
search  
  <input type="text" name="search">
   <select size="1" name="dropdown">
    <option value="" selected>search By...</option>
    <option value="arCompanyname">Company</option>
    <option value="arAddress">Address</option>  
  </select>
  <input type="Submit" value="Submit" name="Submit"> 
</form>

并更改您的查询

SELECT arCompanyname, arAddress FROM ar WHERE $dropdown LIKE '%$search%'

Answer 3

检查更新的代码：

更新了搜索查询mysql_numrows-> mysql_num_rows ，添加了mysql_fetch_array进行循环。

尝试使用mysqli_ *函数编写脚本。 因为mysql_ *函数将不推荐使用。

<html>
    <head>
<basefont face="Arial">
</head>
<body>

<?php
error_reporting(E_ALL); 
if (!isset($_POST['Submit'])) {
// form not submitted
?>

<form action="" method="post">
search  
  <input type="text" name="search">
   <select size="1" name="dropdown">
    <option value="" selected>search By...</option>
    <option value="first">Company</option>
    <option value="last">Address</option>  
  </select>
  <input type="Submit" value="Submit" name="Submit"> 
</form>

<?php
}

else {

    // Server Variables
    $host = "localhost";
    $user = "mdti";
    $pass = "tnet";
    $db = "ojt";

    // Open Connection

    $connect = mysql_connect($host, $user, $pass) or die ("Unable to connect to host");

    $search = empty($_POST['search'])? die ("ERROR: Enter search Criteria") : mysql_escape_string($_POST['search']);
    $dropdown = empty($_POST['dropdown'])? die ("ERROR: Select from dropdown") : mysql_escape_string($_POST['dropdown']);


    //Select Database

    mysql_select_db($db) or die ("Unable to connect to database");

    //Create query

    $query = "SELECT arCompanyname, arAddress FROM ar WHERE $dropdown like '%$search%'";


    $result = mysql_query($query) or die (mysql_error());

    $num=mysql_num_rows($result);

    if($num > 0) {


        echo "<b><center>Database Output</center></b><br><br>";


        while ($row = mysql_fetch_array($result)) {

            $company = $row['arCompanyname'];
            $address = $row['arAddress'];


            echo "<br>Company: $company<br><br>Address: $address<hr><br>";


        }


    } else {

        echo "No rows found";
    }

    mysql_close($connect);  
}
?>
</body>
</html>

Answer 4

使用“ mysql_real_escape_string”

$search = empty($_POST['search'])? die ("ERROR: Enter search Criteria") : mysql_real_escape_string($_POST['search']);
$dropdown = empty($_POST['dropdown'])? die ("ERROR: Select from dropdown") : mysql_real_escape_string($_POST['dropdown']);

并将“ $ search”更改为“％$ search％”

$query = "SELECT arCompanyname, arAddress FROM ar WHERE $dropdown like'%$search%'" or die (mysql_error());

将mysql_numrows更改为mysql_num_rows

用下拉菜单在php中搜索表单

问题描述

4 个解决方案

解决方案1
0 2014-06-11 07:09:20

解决方案2
0 2014-06-11 07:09:57

解决方案3
0 2014-06-11 07:43:47

解决方案4
0 2014-06-11 08:07:55

用下拉菜单在php中搜索表单

问题描述

4 个解决方案

解决方案1 0 2014-06-11 07:09:20

解决方案2 0 2014-06-11 07:09:57

解决方案3 0 2014-06-11 07:43:47

解决方案4 0 2014-06-11 08:07:55

解决方案1
0 2014-06-11 07:09:20

解决方案2
0 2014-06-11 07:09:57

解决方案3
0 2014-06-11 07:43:47

解决方案4
0 2014-06-11 08:07:55