Question

首先，如果这是一个愚蠢的问题我道歉 - 我刚开始接受我的php / mysql技能。我正在制作一个有3个下拉菜单的下拉表格。我希望能够从表单中触发查询。选择零件类型，制作，模型命中提交，并显示结果表。

我的表单中填充了3个数组，当您点击提交时，我可以将每个所选项目的键回显到页面：

    echo '<form action="dbBrowse.php" method="post">';

    $mak = array (0 => 'Make', 'Ford', 'Freightliner', 'Peterbilt', 'Sterling', 'Mack', 'International', 'Kenworth', 'Volvo');
    $mod = array (0 => 'Model', 'Argosy', 'Midliner');
    $p = array (0 => 'Part', 'Radiator', 'Charge Air Cooler', 'AC');                        



    echo '<select name="drop1">';
    foreach ($p as $key => $value) {
    echo "<option value=\"$key\">
    $value</option>\n";

    }


    echo '</select>';




    echo '<select name="drop2">';
    foreach ($mak as $key => $value) {
    echo "<option value=\"$key\">
    $value</option>\n";



    }

    echo '<select/>';



    echo '<select name="drop3">';
    foreach ($mod as $key => $value)  {
    echo "<option value=\"$key\">
    $value</option>\n";



    }

    echo '<select/>';
    echo '</form>';


            //echo keys of selection
    echo $_POST['drop1'];
    echo "<br />";
    echo $_POST['drop2'];
    echo "<br />";
    echo $_POST['drop3'];
            //these echo something like 1, 1, 3 etc. to my page

我迷路的地方是我想要选择选项并将它们插入到这样的查询中：

     $dropSearch = mysql_query('SELECT * FROM parts WHERE part= "$partTypeVar" . AND  WHERE make = "$makeTypeVar" . AND WHERE model = "$modelTypeVar"');    

            $partTypeVar being the corresponding value to the key that is being returned from the array.

我正在疯狂地试图找出如何实现这一目标。最终我想进一步扩展它，但只是能够创建一个带有所选值的mysql语句将使我的一天正好。我理解需要发生什么的概念，但我不确定如何实现它。任何帮助或推动正确的方向将不胜感激。

Answer 1

如果我理解了您的问题，那么在提交form时，您希望query数据库中包含所选值。

使用AND的示例：

// Prepare the Query
$query = sprintf(
           "SELECT * FROM parts WHERE part='%s' AND make='%s' AND model='%s'",
           mysql_real_escape_string($_POST['drop1']),
           mysql_real_escape_string($_POST['drop2']),
           mysql_real_escape_string($_POST['drop3'])
         );

// Execute the Query
mysql_query($query);

这将从表格部分中选择与这三个值匹配的所有行。

使用OR的示例：

// Prepare the Query
$query = sprintf(
           "SELECT * FROM parts WHERE part='%s' OR make='%s' OR model='%s'",
           mysql_real_escape_string($_POST['drop1']),
           mysql_real_escape_string($_POST['drop2']),
           mysql_real_escape_string($_POST['drop3'])
         );

// Execute the Query
mysql_query($query);

这将从表格部分中选择与这三个值中的任何一个匹配的所有行。

您可以阅读更多相关信息：

mysql_query

mysql_real_escape_string

MySQL 5.6 Reference Manual :: 12 Functions and Operators :: 12.3 Operators

Answer 2

首先，你必须删除。在SQL查询中使用char，现在不需要使用它，当然也可以为变量赋值。

$partTypeVar = mysql_real_escape_string($_POST['$drop1']);
$makeTypeVar = mysql_real_escape_string($_POST['$drop2']);
$modelTypeVar = mysql_real_escape_string($_POST['$drop3']);

$dropSearch = mysql_query('SELECT * FROM parts WHERE part= "$partTypeVar" AND WHERE make = "$makeTypeVar" AND WHERE model = "$modelTypeVar"');

我假设你的变量的顺序正确。

我希望这有帮助！

Answer 3

<select name="myFilter">
<option value="Chooseafilter" name="default">Choose a filter...</option>
<option value ="Lastname" name="opLastName">Last Name</option>
<option value="Firstname" name="opFirstName">First Name</option>
<select>
<li><!--TEXT SEARCH INPUT--><input type="text" name="search_filter" /></li> 
...
dbconnection();#assume that all connection data is here        
...
$filter = $_POST['myFilter']; #
...

switch($filter)
    {
        case "Lastname":
        $selectedoption = "profile_name";
        break;

        case "Firstname":
        $selectedoption="profile_first_name";
        break;

        case "Chooseafilter":
        $selectedoption = "";
        break;  
    }           

$result = pg_query($db_con, "SELECT * FROM profile WHERE ".$selectedoption." LIKE'%$_POST[search_filter]%'"); 
$row = pg_fetch_assoc($result);  
if (isset($_POST['submit']))
    {   
        while($row = pg_fetch_assoc($result))
        {
        echo"<ul>  
                <form name='update' action='' method='POST'>
                <li>Guest Last Name:</li>
                <li><input type='text' name='profile_name_result' value='$row[profile_name]' /></li>  
                <li>Guest First Name:</li>
                <li><input type='text' name='profile_first_name_result' value='$row[profile_first_name]' /></li>  
                <li><input type='submit' value='Update' name='update'></input></li>
    ...

查询数据库的下拉菜单

3 个答案: