php的查询功能（如何通过查询条件获取相应表的字段）

比如输入要查询的isbn号，然后确定，在下方输出符合条件的查询结果。

result = mysql_query("select * from books where checked=‘$checked’ ");这个我试了一下，出现以下错误
还有一个问题要请教是如何是要先查询之后才显示表要显示的结果。

... where 字段='$checked' //字段换成你实际的字段

if(isset($_POST['checked'])){
@$result = mysql_query("select * from books",$db);
    echo "<table border=1>\n";
    echo "<tr>
             <td>isbn</td>
             <td>author</td>
             <td>title</td>
             <td>price</td>
             ";
    echo "</tr>\n";
    //循环遍历
      while ($myrow = mysql_fetch_row($result)){
              printf("<tr>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>", $myrow[0], $myrow[1],$myrow[2],$myrow[3]);
            }
         echo "</table>\n";
}

谢谢jordan102，那请问如何通过查询条件，然后在同个页面下输出符合查询条件的表呢？
这是我的表

我试了jordan102的代码，也不知道我添加if(isset($_POST['checked'])){.......}这段代码的位置对不对。
我按了下查询的按钮，但是没反应。代码如下<?php
//获得连接
$db = mysql_connect("localhost", "root", "root") or die(mysql_error());
//echo "Connected to MySQL<br/>";
//连接数据库
mysql_select_db("test") or die(mysql_error());
//echo "Connected to Database";    //查询数据，并用表格显示出来
@$checked = $_POST['checked'];
// @$result = mysql_query("select * from books",$db);
// @$result = mysql_query("select * from books where isbn='$checked' ");
    if(isset($_POST['checked'])){
    @$result = mysql_query("select * from books",$db);
    echo "<table border=1>\n";
    echo "<tr>
             <td>isbn</td>
             <td>author</td>
             <td>title</td>
             <td>price</td>
             ";
    echo "</tr>\n";
    //循环遍历
      while ($myrow = mysql_fetch_row($result)){
              printf("<tr>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>", $myrow[0], $myrow[1],$myrow[2],$myrow[3]);
            }
         echo "</table>\n";
}
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>无标题文档</title>
</head>
<body>
<form action="" name=""><br/>
  <br/>
   <input type="button" value="查询" name="submit"/>
   <input type="text" name="checked"/>
</form>
</body>
</html>

@$result = mysql_query("select * from books where isbn='$checked'",$db);

@jordan102我试了下@$result = mysql_query("select * from books where isbn='$checked'",$db); 但是点了按钮还是不行。

这句<input type="submit" value="查询" name="submit"/>我试了，但是我输入books的isbn到输入框的时候，点击查询，还是查不到isbn=1的对应的全部内容。

$result = mysql_query("select * from books",$db);
条件都不加，还算是条件查询吗？

回复xuzuning，不好意思！忘了加。

回复Jordan120，还是没有表显示。

回复xuzuning，请问这句不是@$result = mysql_query("select * from books where isbn='$checked'",$db);通过查询isbn的编号然后判断符合条件的来输出表。

<?php
//获得连接
$db = mysql_connect("localhost", "root", "root") or die(mysql_error());
//echo "Connected to MySQL<br/>";
//连接数据库
mysql_select_db("test") or die(mysql_error());
//echo "Connected to Database";    //查询数据，并用表格显示出来
// @$checked = $_POST['checked'];
// @$result = mysql_query("select * from books",$db);
// @$result = mysql_query("select * from books where isbn='$checked' ");
    if(isset($_POST['checked'])){
    @$result = mysql_query("select * from books where isbn='$checked'",$db);
    echo "<table border=1>\n";
    echo "<tr>
             <td>isbn</td>
             <td>author</td>
             <td>title</td>
             <td>price</td>
             ";
    echo "</tr>\n";
    //循环遍历
      while ($myrow = mysql_fetch_row($result)){
              printf("<tr>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>", $myrow[0], $myrow[1],$myrow[2],$myrow[3]);
            }
         echo "</table>\n";
}
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>无标题文档</title>
</head>
<body>
<form action="" name="post"><br/>
  <br/>
   <input type="submit" value="查询" name="submit"/>
   <input type="text" name="checked"/>
</form>
</body>
</html>

@$result = mysql_query("select * from books where isbn='$checked'",$db);
在哪里赋值的？？？

嗦嘎，我在#18标红了你还抄错。
<form action="" method="post">

<?php
//获得连接
$db = mysql_connect("localhost", "root", "rDSzPnhnDJCAHYuj") or die(mysql_error());
//echo "Connected to MySQL<br/>";
//连接数据库
mysql_select_db("test") or die(mysql_error());
//echo "Connected to Database";//查询数据，并用表格显示出来
//    @$checked = $_POST['checked'];
//    @$result = mysql_query("select * from books",$db);
//    @$result = mysql_query("select * from books where isbn='$checked' ");
if(isset($_POST['checked'])){
    $checked=$_POST['checked'];
    @$result = mysql_query("select * from books where isbn='$checked'",$db);
    echo "<table border=1>\n";
    echo "<tr>
             <td>isbn</td>
             <td>author</td>
             <td>title</td>
             <td>price</td>
             ";
    echo "</tr>\n";
    //循环遍历
    while ($myrow = mysql_fetch_row($result)){
        printf("<tr>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>", $myrow[0], $myrow[1],$myrow[2],$myrow[3]);
    }
    echo "</table>\n";
}
?>
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>无标题文档</title>
</head>
<body>
<form action="" name="post" method="post"><br/>
    <br/>
    <input type="submit" value="查询" name="submit"/>
    <input type="text" name="checked"/>
</form>
</body>
</html>

<?php
//获得连接
$db = mysql_connect("localhost", "root", "root") or die(mysql_error());
//echo "Connected to MySQL<br/>";
//连接数据库
mysql_select_db("test") or die(mysql_error());
//echo "Connected to Database";//查询数据，并用表格显示出来
//    @$checked = $_POST['checked'];
//    @$result = mysql_query("select * from books",$db);
//    @$result = mysql_query("select * from books where isbn='$checked' ");
if(isset($_POST['checked'])){
    $checked=$_POST['checked'];
    @$result = mysql_query("select * from books where isbn='$checked'",$db);
    echo "<table border=1>\n";
    echo "<tr>
             <td>isbn</td>
             <td>author</td>
             <td>title</td>
             <td>price</td>
             ";
    echo "</tr>\n";
    //循环遍历
    while ($myrow = mysql_fetch_row($result)){
        printf("<tr>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>
                           <td>%s</td>", $myrow[0], $myrow[1],$myrow[2],$myrow[3]);
    }
    echo "</table>\n";
}
?>
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>无标题文档</title>
</head>
<body>
<form action="" name="post" method="post"><br/>
    <br/>
    <input type="submit" value="查询" name="submit"/>
    <input type="text" name="checked"/>
</form>
</body>
</html>

回复qxhaidao、jordan102：折腾了两位一个下午，qxhaidao的代码运行成功，jordan那个只显示了表头。我会给两位加分的，实在谢谢了！

调试易

php的查询功能（如何通过查询条件获取相应表的字段）

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