从动态jQuery表单向MySQL中添加数据 - Adding data into MySQL from a dynamic jQuery form

我正在尝试使用jQuery从动态表单向MySQL插入数据。为了记录在案，我正在使用Twitter Bootstrap我实际上实现了插入"行"，但没有数据我现在找到的最好的帮助是在这里插入MySQL和jQuery脚本。

提前感谢您抽出时间。

文件：add.php

<form  action="../reglas/insert.php" method="post" class="form-horizontal">
<div class="container">
<div id="content">
<div class="row">
<a class="add_line btn btn-small" href="#" style="margin-left:5px">+</a>
</div>
<br>
</div>

</div>
 </div>
 </div>
 <div style="margin:20px auto; width:80px;">
 <input  class="btn btn-success"  type="submit" value="Guardar">
</div>
</form>

<script>
$("#content").delegate(".add_line", "click", function() {
var content = '<div class="row">' +
                     '<div class="span3"><div class="input-append"><input type="text"   class="span2 text-right" placeholder="0.00"  id="enganche"  name="enganche" value="" /><span class="add-on">%</span></div></div>' +
                     '<div class="span3"><div class="input-append"><input type="text" class="span2 text-right" placeholder="0.00" id="comision" name="comision" value="" /><span class="add-on">%</span></div></div>' +
                     '<div class="span3"><div class="input-append"><input type="text" class="span2 text-right" placeholder="0.00" id="r1" name="r1[]" value="" /><span class="add-on">%</span></div></div>' +
                     '<div class="span2"><div class="input-append"><input type="text" class="span1 text-right" placeholder="0.00" id="r2" name="r2[]" value="" /><span class="add-on">%</span></div></div>'+
                     '<a class="del_line btn btn-small" href="#" style="margin-left:5px">-</a>&nbsp;' +
                     '<br><br>'+
                     '</div>';
$("#content").append(content);
return false;   
});
$("#content").delegate(".del_line", "click", function() {
$(this).parent().remove();
return false;  
});
</script>

文件：insert.php

连接：

$con=mysqli_connect("$host","$username","$password","$db_name");
mysqli_autocommit($con, false);
$flag=true;

查询：

$query="
INSERT INTO $tbl_name(
the_input,
)
VALUES (
'$the_input',
)";

"for"：

for ($i = 0; $i < count($_POST['the_input']); $i++) {
$the_input = $_POST['the_input'][$i];

$result = mysqli_query($con, $query);
if (!$result) {
$flag = false;
echo "Error details: " . mysqli_error($con). ". ";
}
}
if ($flag) {
mysqli_commit($con);
echo "Done!";
}
mysqli_close($con);
?>

Okey dokey。

第一件事：

$.delegate()已弃用，除非您使用的版本<jQuery 1.7（它是许多版本以前的），您应该迁移到$.on()事件绑定器/delegator函数。这将确保您的代码经得起未来的考验（至少在他们再次更改api之前！）

第二：

我假设你已经将你的数据包装在一个表单标签中，并且这些值实际上被发送到接收页面，我会这样做：

if(isset($_POST['the_input']))
{
    //create connection object;
    $mysqli = new mysqli($host,$username,$password,$db_name);
    $query = "INSERT INTO `the_input` VALUES (";
    //construct the query. Get the the_input data and paramater string (required for stmt binding)
    foreach($_POST['the_input'] as $value)
    {
        $query .= $value . ",";
    }
    //remove trailing comma
    $query = substr($query, 0, strlen($query)-1);
    $query .= ")";
    //this is just so you can see what the loop did and test the output of the query, remove this when you're confident
    echo "The constructed query string is: " . $query
    if($mysqli->query($query) == false)
    {
        trigger_error($mysqli->error);
    } else {
        echo "Success! Check the DB!";
    }
} else {
    Echo "No POST data was recieved";
}