小编Xav*_*avi的帖子

我想使我的表的标题得到修复.表格存在于可滚动的div中.请在此处查看我的代码:http: //jsfiddle.net/w7Mm8/114/请建议我解决此问题.

谢谢

我的代码:

<div style="position: absolute; height: 200px; overflow: auto; ">
    <div style="height: 250px;">
        <table border="1">
            <th>head1</th>
            <th>head2</th>
            <th>head3</th>
            <th>head4</th>
            <tr>
                <td>row 1, cell 1</td>
                <td>row 1, cell 2</td>
                <td>row 1, cell 2</td>
                <td>row 1, cell 2</td>
            </tr>
            <tr>
                <td>row 2, cell 1</td>
                <td>row 2, cell 2</td>
                <td>row 1, cell 2</td>
                <td>row 1, cell 2</td>
            </tr>
        </table>
    </div>
</div>

我有一个下拉框.当用户从下拉框中选择一个值时,它会执行查询以从数据库中检索数据,并使用ajax在前端显示结果.这需要一点时间,所以在这段时间里,我想展示一个进度条.我已经搜索过了,我已经找到了许多关于创建上传进度条的教程,但我还没有喜欢.有人可以为我提供一些有用的指导吗？

我的ajax代码:

<script>
$(function() {
    $("#client").on("change", function() {
      var clientid=$("#client").val();

        $.ajax({
            type:"post",
            url:"clientnetworkpricelist/yourfile.php",
            data:"title="+clientid,
            success:function(data){
             $("#result").html(data);
            }
        });

    });
});
</script>

我正在尝试显示来自数据库的图像,但我无法显示图像.但它显示为这样user-1.jpg请看我的代码可以指导我如何显示图像.

$sqlimage = "SELECT image FROM userdetail where `id` = $id1";
$imageresult1 = mysql_query($sqlimage);

while($rows = mysql_fetch_assoc($imageresult1))
{       
    $image = $rows['image'];    
    print $image;
}

我有一张表格如下:

    6xx     8xx      9xx     11xx      12xx
  1  0.01    0.002    0.004   0.001     0.025
  2  0.025   0.125    0.002   0.01      0.011

我想从列中找到最小值,使该列为绿色.

例如,在第一个中,最小值是0.001,所以我希望它是绿色,第二个0.002是最小值,我希望它是绿色.

任何人都可以指导我如何做到这一点,谢谢

下面是我如何从数据库中选择它并在表中显示int的代码

<?php

$dbHost = 'localhost'; // usually localhost
$dbUsername = 'xxxx';
$dbPassword = 'xxxx';
$dbDatabase = 'xxxx';
$db = mysql_connect($dbHost, $dbUsername, $dbPassword) or die ("Unable to connect to Database Server.");
mysql_select_db ($dbDatabase, $db) or die ("Could not select database.");



$ColumnNames = mysql_query("SELECT column_name 
                              FROM information_schema.COLUMNS 
                             WHERE table_name = 'supplierprice' 
                               AND column_name NOT IN ('supp_price_id',
                                                       'region',
                                                       'country',
                                                       'net_id',
                                                       'networkname', 
                                                       'mcc', …

我试图在HTML表中显示数据库中的数据,我想在警告框中显示这个html表,现在它在警告框旁边显示数据但是没有显示像html表,现在我的警报框显示为这个:

我想以HTML表格格式显示它.任何人都可以指导我如何做到这一点,

这是我的代码:

阿贾克斯

$(document).ready(function() {

    $("#display").dblclick(function() {                

      $.ajax({    //create an ajax request to load_page.php
        type: "GET",
        url: "supplierprice/retrieve.php",             
        dataType: "html",   //expect html to be returned                
        success: function(response){                    
            $("#responsecontainer").html(response); 
          alert(response);
        }

    });
});
});

HTML

<td><input type="button" id="display" value="" /></td>

Retrieve.php:

include"config.php";
$result=mysql_query("select * from supplierpricehistory");

echo "<table border='1' >
<tr>
<td align=center> <b>supplier</b></td>
<td align=center><b>country</b></td>
<td align=center><b>networkname</b></td>
<td align=center><b>mcc</b></td></td>
<td align=center><b>mnc</b></td>
<td align=center><b>newprice</b></td>       
<td align=center><b>oldprice</b></td>       
<td align=center><b>status</b></td>     
<td align=center><b>date</b></td>           
<td align=center><b>user</b></td>";

while($data = mysql_fetch_row($result))
{   
    echo "<tr>";
    echo …

我想这样输出

var data = { 
{
"cname": "Albania"
}
{
"cname": "Austria"
}
}

这个代码我试过,但我收到一个错误 "Parse error: syntax error, unexpected T_VAR in C:\wamp\www\mvc\map\data.php on line 22"

<?php
header('Content-type: application/json');  // this is the magic that sets responseJSON

 $dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "test";

// Connecting, selecting database
$link = mysql_connect($dbhost, $dbuser, $dbpass)
    or die('Could not connect: ' . mysql_error());
mysql_select_db($dbname) or die('Could not select database');

        $query = "SELECT cname FROM country";       // Performing SQL …

我在Excel中有A和B列,如果数字在两列中,我想在列C中自动写入"存在",在值B列中出现的行中.

例:

A   |  B | C 
1   |  2 | Not Exist
3   |  4 | Not Exist
5   |  1 | Exist
6   |  7 | Not Exist

我从数据库中获取手机号码我有更多的1000手机号码,我的数据库看起来像这样

   mobilenumber
   971525478965
   919844005522
   45712345678

我想通过数据库中的每个数字,并countrycode从手机号码显示countrycode和使用PHP的国家

比如这样

 countrycode 971 country UAE
 countrycode 91 country India
 countrycode 45 country Denmark

任何人有任何建议请指导我如何做到这一点.

我试过这样,但想从数据库中检查多个移动数据

 <?php

$number = "971527139011";

    $countrys = array(

    '1' => 'us',
    '2' => 'uk',
    '3' => 'de',
    '44' => 'fi',
    '123' => 'no',
    '971' =>'uae',
    '91' =>'india',
    '92' =>'pakistan'
    );

    $i = 4;
    $country = "";
    while ($i > 0) {
        if (isset($countrys[substr($number, 0, $i)])) {
            $country = $countrys[substr($number, 0, $i)];
            break;
        } else {
            $i--; …

我正在尝试从联系表单发送电子邮件,因此发送电子邮件正在工作但发送后我想重新定向到contact_formsuccess.php,但它没有重定向到那个页面,它仍然是相同的,可以任何一个如何谢谢你

<?php
$target_path = "/var/www/xxxxxx/cv/";

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 

//print $target_path; 

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name']). 
    " has been uploaded";
} else{
    echo "There was an error uploading the file, please try again!";
}


$name=$_POST['senderName'];
$email=$_POST['senderEmail'];
$mobile=$_POST['senderMobile'];
$company=$_POST['senderCompany'];
$inquiry=$_POST['inquiry'];
$message=$_POST['message'];

include 'class.phpmailer.php';
include 'class.smtp.php';

$mail = new PHPMailer(); // create a new object
$mail->IsSMTP(); // enable SMTP
$mail->SMTPDebug = 1; // debugging: 1 = errors and messages, 2 = messages only
$mail->SMTPAuth = true; …