小编Jak*_*Ols的帖子

使用JavaScript显示/隐藏'div'

对于我正在做的网站,我想加载一个div,然后隐藏另一个div,然后有两个按钮,使用JavaScript在div之间切换视图.

这是我目前的代码

function replaceContentInContainer(target, source) {
  document.getElementById(target).innerHTML = document.getElementById(source).innerHTML;
}

function replaceContentInOtherContainer(replace_target, source) {
  document.getElementById(replace_target).innerHTML = document.getElementById(source).innerHTML;
}

Run Code Online (Sandbox Code Playgroud)

<html>
<button onClick="replaceContentInContainer('target', 'replace_target')">View Portfolio</button>
<button onClick="replaceContentInOtherContainer('replace_target', 'target')">View Results</button>

<div>
  <span id="target">div1</span>
</div>

<div style="display:none">
  <span id="replace_target">div2</span>
</div>

Run Code Online (Sandbox Code Playgroud)

替换div2的第二个函数不起作用,但第一个是.

html javascript onclick

Jak*_*Ols

2019 01-23

171
推荐指数

8
解决办法

106万
查看次数

插入MySQL表PHP

我在制作一个简单的表单以将数据插入MySQL表时遇到了一些麻烦.我一直收到这个SQL错误:

"错误:您的SQL语法中有错误;请查看与您的MySQL服务器版本对应的手册,以便在'stock('ItemNumber','Stock')附近使用正确的语法VALUES('#4','3' ')'在第1行"

我的表格HTML是:

    <form action="database.php" method="post">
    Item Number: <input type="text" name="ItemNumber">
    Stock: <input type="text" name="Stock">
    <input type="submit">
    </form>

Run Code Online (Sandbox Code Playgroud)

而PHP是:

    <?php
    $con=mysqli_connect("localhost","root","root","inventory");
    if (mysqli_connect_errno($con))
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
     $sql = "INSERT INTO current stock ('ItemNumber', 'Stock')
    VALUES
    ('$_POST[ItemNumber]','$_POST[Stock]'')";
    if (!mysqli_query($con,$sql))
      {
      die('Error: ' . mysqli_error($con));
      }
    echo "1 record added";
    mysqli_close($con);
    ?>

Run Code Online (Sandbox Code Playgroud)

php mysql sql syntax-error sql-insert

Jak*_*Ols

2014 07-01

4
推荐指数

1
解决办法

6万
查看次数

Flask 405错误

尝试将一些简单文本插入数据库时,我不断收到此错误.

Method Not Allowed
The method is not allowed for the requested URL."

Run Code Online (Sandbox Code Playgroud)

我正在从PHP转向python,所以请耐心等待.

代码是:

from flask import Flask, request, session, g, redirect, url_for, \
    abort, render_template, flash

from flask.ext.sqlalchemy import SQLAlchemy

app = Flask(__name__)
    app.config['SQLALCHEMY_DATABASE_URI'] = 'mysql://root:password@localhost/pythontest'
    db = SQLAlchemy(app)
    app = Flask(__name__)

@app.route('/justadded/')
def justadded():
    cur = g.db.execute('select TerminalError, TerminalSolution from Submissions order by id desc')
    entries = [dict(title=row[0], text=row[1]) for row in cur.fetchall()]
    return render_template('view_all.html', entries=entries)

@app.route('/new', methods= "POST")
def newsolution():
    if not request.method == 'POST':
        abort(401) …

Run Code Online (Sandbox Code Playgroud)

python flask web

Jak*_*Ols

2013 07-30

0
推荐指数

1
解决办法

2470
查看次数