小编Der*_*376的帖子

我正在开发一个cms项目,它与bootstrap WYSIWYG表单有关,用于从数据库中插入和检索.插入代码正常工作,检索代码也可以正常工作,但在我想编辑文章时不起作用.当我点击编辑链接时,<a href='index.php?page=edit&id=".$row['id']."'><span data-placement='top' data-toggle='tooltip' title='Edit'><button class='btn btn-primary btn-xs' data-title='Edit' ><span class='glyphicon glyphicon-pencil'></span></button><span></a>它会将我引导到我的编辑页面.在我的edit.php页面上,我有这个代码从数据库中选择,它运行良好

<?php
include("dbconnect.php");

if(isset($_GET['id']))

$id = strip_tags($_GET['id']);
$sql = "SELECT * FROM berita WHERE id=$id" ;
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($result))  
{
    $image= $row['gambar'];
    $title = $row['judul'];
    $description = ( $row['konten']);
    $time = $row['tanggal'];
}
?>

当我将值回显到它们各自的表单类型时,它仅适用于基于Bootstrap的WYSIWYG不回显任何值,但如果我将其更改为普通textarea,它可以正常工作.这是我在edit.php页面上的代码

<?php
include("dbconnect.php");


if(isset($_GET['id']))    
    $id = strip_tags($_GET['id']);

$sql = "SELECT * FROM berita WHERE id=$id";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($result))      
{
    $image= $row['gambar']; …

我试图插入两个表,但得到这个错误

错误:INSERT INTO provide_help(amount)VALUES(40,000.00)列数与第1行的值计数不匹配

下面是我的插入代码

<?php

    session_start(); {



    //Include database connection details
    include('../../dbconnect.php');


$amount =  strip_tags($_POST['cat']);
$field1amount = $_POST['cat'];
$field2amount = $field1amount + ($field1amount*0.5);



$sql = "INSERT INTO provide_help (amount) VALUES ( $field1amount)";
if (mysqli_query($conn, $sql)) 


$sql = "INSERT INTO gh (ph_id, amount) VALUES (LAST_INSERT_ID(), $field2amount)";
if (mysqli_query($conn, $sql)) 

 {
    $_SESSION['ph'] ="<center><div class='alert alert-success' role='alert'>Request Accepted.</div></center>";
   header("location: PH.php");
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
}
?>

但是,当我做这样的事情时,它有效

$sql = "INSERT INTO provide_help …