在 PHP 中显示表格

Question

我必须显示来自 mysqli 数据库的大量数据列表，但是当我试图显示数据时，表中只显示第一项（这里有多个数据来自单个 id，大约有 16,000 个 id ，就像一个 id 即站点它们是门警报、站点警报、电池警报和故障警报）。我试过 while 循环，但它在这里不起作用。

代码

<?php


include 'dbinc.php';

mysql_connect($mysql_hostname,$mysql_user,$mysql_password);//database connection
mysql_select_db($mysql_database);
                
$order = "SELECT * FROM quanta
ORDER BY ID DESC 
                LIMIT 1
  ";
//order to search data
//declare in the order variable
                
$result = mysql_query($order);  
//order executes the result is saved
//in the variable of $result



    $data = mysql_fetch_row($result);
    
    
    $data_id        = $data[0];     //Unique Id
    $data_timestamp = $data[1];     //Time Stamp by Server
    $data_validated = $data[2];     //Validated by Server
    
    $data_type      = $data[3];     //Type 0 - Periodic / 1 - Fault
    $data_type_str = "Unknown";
    if ($data[3] == 0) $data_type_str = "Periodic";
    if ($data[3] == 1) $data_type_str = "Fault";
    
    $data_site_id   = $data[4];     //String Site Id
    $data_datetime  = $data[5];     //Date Time by Device
    $data_device_id = $data[6];     //Device Id/ Only 1 known for now
    $data_status    = $data[7];     //Status Bits - To Expand
    $data_status_hex= bin2hex($data_status);
    
    $data_raw       = $data[8];     //Raw Data, for device specific information
    
    
    echo ("<table class='table table-striped'>");
    
    echo("
            <tr><td>Site Id</td><td>$data[4]</td></tr>
            <tr><td>Date Time</td><td>$data_datetime</td></tr>
            <tr><td>Device Id</td><td>$data_device_id</td></tr>");
    
    //echo ("Status: <b>$data_status</b><br>");
    //echo ("Status: <b>$data_status_hex</b></tr>");
    
    //$tmp = gettype($data_status);
    //echo "Var: $tmp<br>";
    
    $stats = unpack ( "C*" , $data_status );
    //var_dump($stats);
    



    //for table









      
    
    echo "<tr><td>Smoke Fire Alarm</td><td";
    //0x0000000000 Smoke fire    0 means No alarm,  1 means Alarm
    if (($stats[5] & 0x01) == false) 
        echo " >no Smoke    alarm"; 
    else
        echo " >smoke alarm";
    
    echo "</td></tr>" ;
    
    echo "<tr><td>Door</td><td";
    //$FLAG_01_DOOR     = 0x0000000002; //Door Open     0  means  Door Close , 1 Means  Door open  
    if (($stats[5] & 0x02) == false) 
        echo "  >door closed";
    else
        echo "  >door open";

    echo "</td></tr>";
    
    
    echo "<tr><td>Mode</td><td";
    //$FLAG_02_AUTO     = 0x0000000004; //Auto/Man Mode 0 Means  Auto Mode, 1 Means  Man Mode
    if (($stats[5] & 0x04) == false) 
        echo "  >Auto";
    else
        echo "  >Manual";
    echo "</td></tr>";
    
    echo "<tr><td>Load</td><td>";
    //$FLAG_034_LOAD    = 0x0000000018; //00    :Load on EB,01: Load on DG, 10: Load on site Battery   11: Not used
    if (($stats[5] & 0x08) == false) 
    {
        if (($stats[5] & 0x10) == false) 
            echo "[- Load on EB -]<br>";                //00
        else
            echo "[- Load on site Battery -]<br>";  //10
    }
    else
    {
        if (($stats[5] & 0x10) == false) 
            echo "[- Load on DG -]<br>";                //01
        else
            echo "[- Not Used -]<br>";  //11
    }
    echo "</td></tr>";




    echo "<tr><td>Alternate Fault</td><td";
    //$FLAG_10_ALT  = 0x0000000400; //Alternate Fault
    if (($stats[4] & 0x04) == false) 
        echo "  >no fault";
    else
        echo "  >fault";
    echo "</td></tr>";


   





    
?>

Answer 1

它只显示 1 个项目，因为您只执行 1 个 mysql_fetch_row($result)。 也许试试这个

while($data = mysql_fetch_row($result)) {
   //The code you have after to generate the table
}

文档：https://www.php.net/manual/en/function.mysql-fetch-row.php

Answer 2

如果您希望显示所有项目，则需要使用 while 循环

while($data = mysql_fetch_row($result)){

//在此处输入您的代码

}