我正在尝试做下一件事:
从表格电影中选择行标题,并显示<h3></h3>标签之间的答案.
这就是我现在拥有的.
$sql = "SELECT title FROM movies";
$result = mysqli_query($mysqli, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result))
?>
<div class="grids">
<div class="grid">
<h3>
<?php {
echo "" . $row["title"]. "";
}
}?>
</h3>
2个div类正在代码中的某个地方结束.当我在h3标签之间添加纯文本时,它可以很好地工作.
这也很完美,但它并没有在我希望它的地方回应:
<?php
$sql = "SELECT title FROM movies";
$result = mysqli_query($mysqli, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "" . $row["title"]. " <br>";
} …我试图通过添加已存在的原始数字值来更新我的数据库中的字段.我有一个系统,员工可以登录并更新普通用户的余额.目前我有一个测试用户和工作人员.用户余额设置为100.我有以下代码:
<?php
if(isset($_POST['search'])){
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$result = $mysqli->query( "SELECT * FROM Users WHERE Username ='$searchq'");
if ($result){
//fetch result set as object and output HTML
if($obj = $result->fetch_object())
{
echo '<div class="booksearched">';
echo '<form method="POST" id = "books" action="">';
echo '<div class="book-content"><h3>Student Username: '.$obj->Username.'</h3>';
echo '<br>';
echo '<div class="book-content"><i>First Name: <b>'.$obj->FirstName.'</b></i></div>';
echo '<div class="book-desc"><i>Last Name:<b> '.$obj->LastName.'</b></i></div>';
echo '<br>';
echo '<div class="book-qty"> Current Balance<b> '.$obj->Balance.'</b></div>';
echo 'New Balance: <input type="number" name="newBalance" value = "1" min = "1" …我昨天开始学习PDO,我以为我把它弄下来但是在尝试提交一个简单的表格时我又遇到了一个错误.我在w3schools上使用这个例子,但是有一个表单输入.
的index.php
<form action="submit.php" method="post">
<input type="text" id="name" placeholder"Enter Your Name">
<button type="submit">Submit</button>
</form>
这是我处理表单的页面:
submit.php
<?php
$servername = "localhost";
$username = "testuser";
$password = "testpassword";
$dbname = "testdb";
$nickname = $_POST['name'];
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
//Set PDO Error Mode to Exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "INSERT INTO test (name)
VALUES (:nickname)";
// use exec() because no results are returned
$conn->exec($sql);
echo "New record created successfully";
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage(); …我正在为一个招聘机构制作一个原型网站,我在将任何数据上传到为这个原型系统创建的数据库时遇到了问题.这是注册新用户页面,它不会将html文本字段中的任何信息发送到数据库.
这是我的代码 -
<?php
require (__DIR__.'/connections/connections.php');
if(isset($_POST['Register'])){
session_start();
$Title = $_POST['Consultant_Title'];
$Fname = $_POST['Consultant_First_Name'];
$Lname = $_POST['Consultant_Last_Name'];
$Cbranch = $_POST['Consultant_Branch'];
$Email = $_POST['Email'];
$Username = $_POST['Username'];
$Password = $_POST['Password'];
$StorePassword = password_hash($Password, PASSWORD_BCRYPT, array('cost' => 10));
$spl = $con->query("INSERT INTO consultant_details (Title, Fname, Lname, Cbranch )Values('{$Title}', '{$Fname}', '{$Lname}', '{$Cbranch}')");
$spl = $con->query("INSERT INTO users ( Email, Username, Password )Values('{$Title}', '{$Email}', '{$Username}', '{$StorePassword}')");
header('location: Login.php');
}
?>
Html -
<h2>Staff Register </h2>
<div class="clr"></div>
<form action="" method="post" name="Registerform" id="Registerform">
<div class="FormElement"><select …目前我想在用户登录系统时从两个表中获取两个不同的数据.权利表是"user","user_staff"和"user_group".但是当用户输入其用户名和密码并提交它时,提示"致命错误:在第26行的C:\ xampp\htdocs\auditsystem\index.php中的非对象上调用成员函数fetch_array()"
以下是代码:
if($username!= "" && $password != "")
{
//INNER JOIN user_group_module_role ON user.user_group_module_role_id = user_group_module_role.id
$result = $db->query("SELECT * FROM user
INNER JOIN user_staff ON user.user_staff_id = user_staff.id
WHERE username = '$username' AND password = '$password'");
if($result->num_rows == 1)
{
$validate = $result->fetch_assoc();
$query1 = "SELECT * FROM usergroup WHERE id = $validate[user_group_id]";
$result1 = $db->query($query1);
$row1 = $result1->fetch_array();
//change here for the authority
$_SESSION['user_staff'] = $validate['displayname'];
$_SESSION['usergroup'] = $row1['user_group_type'];
echo "<script language='javascript'>window.location='panel.php'</script>";
}
else
{
echo …我用PHP和MySQL数据库创建了票务系统.
在视图故障单页面上,我显示了数据库中的每个故障单,其中包括每个故障单的优先级.优先级设置为低,正常或高.
当数据传递到我的页面时,它会将div中的优先级设置为我在数据库表中所说的内容.
我想找到一种方法来改变背景颜色,具体取决于优先级字段中显示的值,所以红色表示高,橙色表示正常,绿色表示低.
HTML/PHP
<div class="priority">
<?php
$priority_query = "SELECT priority FROM tickets";
$priority = mysql_query($priority_query);
if ($priority = 'low') { ?>
<div class="priority_low"><?php echo 'Priority: ' . mysql_result ($result,$j,'priority') .''; ?></div>
<?php
} else if ($priority = 'normal') { ?>
<div class="priority_normal"><?php echo 'Priority: ' . mysql_result($result,$j,'priority') .''; ?></div>
<?php
} else if ($priority = 'high') { ?>
<div class="priority_high"><?php echo 'Priority: ' . mysql_result($result,$j,'priority') .''; ?></div>
<?php } ?>
</div>
CSS
.priority_low {
background-color: green;
}
.priority_normal …我有这个代码但我面临铬的错误,没有定义load_new_content_dr ...
<html>
<head>
<script lang="javascript" src="http://code.jquery.com/jquery-latest.js">
$(document).ready(function(){
$("#Dr_name").change(load_new_content_dr());
$("#day").change(load_new_content_day());
});
function load_new_content_dr(){
var selected_option_value=$("#Dr_name option:selected").val();
$.post("_add_reservasion.php", {option_value: selected_option_value},
function(data){
$("#day").html(data);
alert(data);
}
);
$.post("__add_reservasion.php",
function(data){
$("#time").html(data);
alert(data);
}
);
}
function load_new_content_day(){
var selected_day = $("#day option:selected").val();
$.post("__add_reservasion.php", {selected_day:selected_day},
function(data){
$("#time").html(data);
alert(data);
}
);
}
</script>
</head>
<body>
...
some codes for connecting database and other stiffs...
...
<!--======================doctore name=========================-->
<p>doctor name:</p>
<select name="Dr_name" id = "Dr_name" form="new" onchange="load_new_content_dr()">
<option value="null"></option>
<?php
while($row = mysqli_fetch_array($table))
{
$name = $row["name"]; …我想弄清楚mysql -u root -p命令的作用。
我用谷歌搜索了命令，但找不到任何好的结果。
我试图在大写字母后找到所有数字.请参阅以下示例:
E1S1 应该给我一个数组包含: [1 , 1]
S123455D1223 应该给我一个数组包含: [123455 , 1223]
我试过以下但没有得到上面显示的任何例子的任何匹配:(
$loc = "E123S5";
$locs = array();
preg_match('/\[A-Z]([0-9])/', $loc, $locs);
任何帮助非常感谢我是regex的新手.
这是我以前要检查的代码是否$A不匹配$B
if($A!=$B) {
$set = array();
echo $val= str_replace('\\/', '/', json_encode($set));
//echo print_r($_SERVER);
exit;
}
现在我需要与这种情况相反:( $A需要匹配其中一个$B,$C或者$D)