0

插入基于搜索结果的多条记录时出现问题。搜索字段工作正常。但问题是插入不起作用。此外，在我插入记录之前，我必须编辑任何我想要的“等级”字段部分。这是我的代码。

<?php
require_once('../../_includes/initialize.php');
?>


<html><br />

<form action="search.php" method="POST">
<?php
echo "<b><center>Search Results</center></b><br><br>";

if (isset($_POST['search_name'])) {

    $search_name = $_POST['search_name'];
if (!empty($search_name)) { 

$query="SELECT * FROM students WHERE course like                          '%".mysql_real_escape_string($search_name)."%'";

$result=mysql_query($query);
$num=mysql_num_rows($result);

if (($num) >=1) {
//echo 'Found.';

$i=0;
while ($i < $num) {
$id=mysql_result($result,$i,"id");
$f_name=mysql_result($result,$i,"f_name");
$g=mysql_result($result,$i,"dump");

?> 

<input name="id" type="text" value="<?php echo @$id; ?>" size="40" />
<input name="f_name" type="text" value="<?php echo @$f_name; ?>" size="40" />
<input name="a" type="text"  value="<?php echo $g; ?>" size="40" />

<br />

<?php
$i++;
}   

} else {

echo 'No results found.';
}  
}
}

?>
<?php
if (isset($_POST['Submit_Grade'])) {

$query = "INSERT INTO samplestudgrade (id, name, grade) VALUES ('".$_POST["id"]."','".$_POST["f_name"]."','".$_POST["a"]."')";
            $query_run = mysql_query($query);

            }

?>

<input name="search_name" type="text" value="search..." />
<input type="submit" value="Search"/>
<input type="submit" name="Submit_Grade" value="submit" />
</form>
</html>
4

1 回答 1

0

试试这个代码：

编辑：

<html><br />

    <form action="search.php" method="POST">
        <?php
        echo "<b><center>Search Results</center></b><br><br>";

        if (isset($_POST['search_name'])) {

            $search_name = $_POST['search_name'];
            if (!empty($search_name)) {

                $query = "SELECT * FROM students WHERE course like '%" . mysql_real_escape_string($search_name) . "%'";

                $result = mysql_query($query);
                $num = mysql_num_rows($result);

                if (($num) >= 1) {
                    //echo 'Found.';

                    $i = 0;
                    while ($i < $num) {
                        $id = mysql_result($result, $i, "id");
                        $f_name = mysql_result($result, $i, "f_name");
                        $g = mysql_result($result, $i, "dump");
                        ?> 

                        <input name="id[]" type="text" value="<?php echo @$id; ?>" size="40" />
                        <input name="f_name[]" type="text" value="<?php echo @$f_name; ?>" size="40" />
                        <input name="a[]" type="text"  value="<?php echo $g; ?>" size="40" />

                        <br />

                        <?php
                        $i++;
                    }
                } else {

                    echo 'No results found.';
                }
            }
        }
        ?>
        <?php
        if (isset($_POST['Submit_Grade'])) {
            $ctr = 0;
            foreach ($_POST['id'] as $id) :
                if ($_POST["a"] >= 75) {
                    $status = "Passed";
                } elseif ($_POST["a"] >= 73) {
                    $status = "Conditional";
                } elseif ($_POST["a"] >= 70) {
                    $status = "Failed";
                } else {
                    $status = "Incomplete";
                }
                $query = "INSERT INTO samplestudgrade (id, name, grade, status) VALUES ('" . $id . "','" . $_POST["f_name"][$ctr] . "','" . $_POST["a"][$ctr] . "','" . $status . "')";
                $query_run = mysql_query($query);
                $ctr++;
            endforeach;
        }
        ?>

        <input name="search_name" type="text" value="search..." />
        <input type="submit" value="Search"/>
        <input type="submit" name="Submit_Grade" value="submit" />
    </form>
</html>
于 2012-11-05T08:19:51.553 回答