在我的HTML代码中,我有一个表但最初没有加载数据行,只有标题和列标签。单击按钮后,将根据SQL语句中从数据库中选择的行数将多个行加载到此表中。在这种情况下,如果我的团队在数据库中有七个注册的玩家,我的表将显示七行(通过PHP echo语句)。在每一行中,我都有一个包含玩家姓名和文本输入框的复选框。
这是在点击结果后按钮后添加行的jQuery:
$("#post-result").click(function(){
$("#post-result").hide();
$("#confirm-result").show();
$("#cancel").show();
$("#home-player-count").show();
$("#home-score").html("<input class='form-input f-score' id='home-score-input' type=number name=home_score maxlength=2>");
$("#away-score").html("<input class='form-input f-score' id='away-score-input' type=number name=away_score maxlength=2>");
$("#stats-home").after(<?php
echo "\"";
$i = 0;
$sql2 = mysqli_query($link, "SELECT * FROM user_accounts WHERE ps4_club= '" . $myclub . "'");
while ($row = mysqli_fetch_array($sql2, MYSQLI_ASSOC))
{
$i++;
echo "<tr id='player-".$i."'><td class='match-stats-name'><input type='checkbox' class='f-player' checked='true' value='" . $row['username'] . "' name='home_player_".$i."'>". $row['username'] ."";
echo "<td><input class='form-input f-stats single' maxlength=1 name='home_player_".$i."_rating_1'> . <input class='form-input f-stats single' maxlength=1 name='home_player_".$i."_rating_2'></td>";
echo "<td><input class='form-input f-stats single' maxlength=1 name='home_player_".$i."_goals'></td>";
echo "<td><input class='form-input f-stats single' maxlength=1 name='home_player_".$i."_assists'></td>";
echo "<td><input class='form-input f-stats' maxlength=2 name='home_player_".$i."_tackles_won'> / <input class='form-input f-stats' maxlength=2 name='home_player_".$i."_tackles_made'></td></tr>";
}
echo "\"";
?>)
<?php
$_SESSION["match_id"] = $id;
$_SESSION["player_count"] = $i;
$_SESSION["team"] = $team;
?>
$(&#34; #stats-home&#34;)。之后是添加行的地方。
我遇到的问题是我试图将这些数据从表格输入到我的数据库中。这是我点击表单的确认结果按钮后加载的PHP代码:
<?php
include "../config.php";
if (isset($_POST["confirm_result"]))
{
session_start();
for ($i = 1; $i <= $_SESSION["player_count"]; $i++)
{
$sql = $link->prepare("INSERT INTO ps4_apl_1_stats (match_id, team, name, rating, goals, assists, tackles_won, tackles_made)
VALUES (?, ?, ?, ?, ?, ?, ?, ?)");
$sql->bind_param("issiiiii", $match_id, $team, $name, $rating, $goals, $assists, $tackles_won, $tackles_made);
$match_id = $_SESSION["match_id"];
$team = mysqli_real_escape_string($link, $_SESSION["team"]);
$name = "test_".$i."_player";
$rating = $_POST["home_player_".$i."_rating_1"];
$goals = $_POST["home_player_".$i."_goals"];
$assists = $_POST["home_player_".$i."_assists"];
$tackles_won = $_POST["home_player_".$i."_tackles_won"];
$tackles_made = $_POST["home_player_".$i."_tackles_made"];
$sql->execute();
$sql->free_result();
}
header("Location: ../match.php?id=".$_SESSION['match_id']);
}
else
{
header("Location: ../match.php?id=".$_SESSION['match_id']);
}
?>
这段代码的作用是通过递增$ i变量来循环遍历表中的每一行。在每次迭代中,它将输入框中的数据放入变量中,然后将变量放入准备好的SQL语句中。每行的输入框都有名称,这些名称由它们的行号编号,例如:
第1行:home_player_1_goals,home_player_1_assists
第2行:home_player_2_goals,home_player_2_assists
第3行:home_player_3_goals,home_player_3_assists
当我尝试提交此表单时,我收到了很多&#34; Unidentified index&#34;通知,像这样:
注意:未定义的索引:第20行的C:\ Apache24 \ htdocs \ script \ match_post.php中的home_player_1_goals
由于某种原因,它没有找到通过echo语句动态插入页面的任何输入框的POST变量。找到$ _POST [&#34; confirm_result&#34;]是没有问题的,因为该输入框在开头就被加载到页面中。
我如何解决这个问题?它几乎无法识别通过echo语句加载的任何输入框。
我还使用print_r:
转储$ SESSION和$ POST变量print_r($ _ SESSION):数组([email] =&gt; my_email@hotmail.com [userid] =&gt; 43 [xbox_club] =&gt; [ps4_club] =&gt; Wrecking Crew [match_id] =&gt; 3 [player_count] =&gt; 7 [团队] =&gt;主页)
print_r($ _ POST):数组([confirm_result] =&gt;确认结果)
它只显示最初加载到页面上的confirm_result按钮。
我还有另一个问题:
我如何根据是否检查来跳过插入行?在每一行中我都有一个复选框。禁用此复选框后,将禁用所有行输入框。跳过这一行的最佳方法是什么?目前,无论输入框是否被禁用,我的代码都会遍历每一行并输入数据。
提前致谢!
UPDATE - 这是在&#34;后期结果&#34;之后生成的单行。单击按钮:
<tr id="player-1">
<td class="match-stats-name"><input type="checkbox" name="home_player_1" value="Syrian2nv" checked="true" class="f-player">Syrian2nv</td>
<td><input name="home_player_1_rating_1" maxlength="1" class="form-input f-stats single"> . <input name="home_player_1_rating_2" maxlength="1" class="form-input f-stats single"></td>
<td><input name="home_player_1_goals" maxlength="1" class="form-input f-stats single"></td>
<td><input name="home_player_1_assists" maxlength="1" class="form-input f-stats single"></td>
<td><input name="home_player_1_tackles_won" maxlength="2" class="form-input f-stats"> / <input name="home_player_1_tackles_made" maxlength="2" class="form-input f-stats"></td>
</tr>
答案 0 :(得分:0)
与#stats-home
相关的表单在哪里?你告诉jQuery在之后添加那些表单输入,所以如果你的表单在里面,他们就会错过表单。
如果我是你,我会通过在正常加载页面时输出你的字段来简化整个jQuery调用,但是将一个类应用到你将告诉它不可见的表行:
echo "<tr class='player-information' id='player-".$i."'><td class='match-stats-name'><input type='checkbox' class='f-player' checked='true' value='" . $row['username'] . "' name='home_player_".$i."'>". $row['username'] ."";
<style type="text/css">
.player-information {
display: none;
}
</style>
然后在您的jQuery中,将$("#stats-home").after()
替换为:
$('tr.player-information').show();
不确定这是否能解决您的问题,但至少您会知道您的表单元素在DOM中并在正确的位置输出(ps如果这没有帮助,请发布结果页面上的HTML输出点击按钮后,以便我们看到正在发生的事情)
- 编辑 - 也,发布加载页面时生成的javascript,即PHP完成后的事情
答案 1 :(得分:0)
我会回答第二个问题。就第一个问题而言,基于$ _POST上的var_dump为空的事实,我认为在jQuery插入之后html并不完全正确。
第二个问题,如果未选中复选框,则跳过一行。所以如果你的HTML是这样的:
<tr>
<td><input type="checkbox" name="checkme[]"></td>
<td><input type="text" name = "homePlayerRating[]"></td>
<td><input type="text" name="homePlayerGoals[]"></td>
</tr>
<tr>
<td><input type="checkbox" name="checkme[]"></td>
<td><input type="text" name = "homePlayerRating[]"></td>
<td><input type="text" name="homePlayerGoals[]"></td>
</tr>
然后在php处理它时:
foreach($_POST['homePlayerRating'] as $key=>$homePlayerRating){
if(isset($_POST['checkme'][$key])){
//now you know that row was checked
$playerRating = $_POST['homePlayerRating'][$key];
$goals = $_POST['homePlayerGoals'][$key];
}
}
编辑:这是错误的。我不确定我在想什么。 $_POST['checkme']
数组与该行的其余部分不匹配。如果未选中,则不会将其添加到$ _POST数组中。例如,如果您选中
$_POST['homePlayerTating']
键将是2但
$_POST['checkme']
键将为零,因为它未添加到该数组,除非它已被选中,因此它将是$_POST['checkme']
数组中的第一项。