如何使用php解析信息從mysql數據庫到html頁面

Question

我創建了一個HTML頁面，它接受來自用戶的輸入，我需要獲取特定行的信息。

下面是保存在 「ProcessDetails.html」

<form action="details.php" method="get"/> 
<h3 align="center"><FONT color=#CCFF66>ENTER SAMPLE NAME</h3> 
<p align="center"> 

<input type="text" id="Samplename" name="Sample_name"/> 

</p> 
<div style="text-align:center"> 
<button  type="submit" value="SEARCH"> 
    <img alt="ok" src= 
    "http://www.blueprintcss.org/blueprint/plugins/buttons/icons/tick.png" /> 
    SEARCH 
    </button> 
</form> 

下面是保存爲 「details.php」

<?php 
$userinput = $_GET['Sample_name']; 
    $servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "ProcessTrackingSystem"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $dbname); 
// Check connection 
if ($conn->connect_errno) { 
printf("Connect failed: %s\n", $conn->connect_error); 
exit(); 
} 

$result = mysqli_query($conn, "SELECT * FROM ProcessDetails WHERE Sample_name = '$userinput'") or die(mysqli_error($conn)); 

$row = mysqli_fetch_assoc($result); 



while ($row=mysqli_fetch_row($result)) 
{ 
printf ("%s (%s)\n",$row[0],$row[1]); 
} 



#printf ("SO_Number: %s \n",$row["SO_Number"]) 
#print_r($row); 

printf ("SO_ID:->"); 
printf ($row['SO_ID']); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("No of samples:->"); 
printf ($row['No_of_samples']); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Sample name:->"); 
printf ($row['Sample_name']); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Client name:->"); 
printf ($row["Clientname"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Institution:->"); 
printf ($row["Institution"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Run number:->"); 
printf ($row["Runnumber"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Obtained reads:->"); 
printf ($row["Obtainedreads"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Rerun Info:->"); 
printf ($row["RerunInfo"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Total reads:->"); 
printf ($row["Totalreads"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Run date:->"); 
printf ($row["Rundate"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Raw data location:->"); 
printf ($row["Rawdatalocation"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Analyst:->"); 
printf ($row["Analyst"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Mentor:->"); 
printf ($row["Mentor"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Analysis start date:->"); 
printf ($row["Analysisstartdate"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Analysis end date->:"); 
printf ($row["Analysisenddate"]); 
printf ("<br>\r\n"); 
printf ("<br>\r\n"); 

printf ("Report location->:"); 
printf ($row["Reportlocation"]); 



mysqli_free_result($result); 
$conn->close(); 
?> 

php腳本的HTML代碼，我需要在表格格式的所有這些數據基於Sample name行。 現在它不會在新的網頁中顯示任何輸出。

幫助我這樣做，在此先感謝。

Answer 1

刪除此行

$row = mysqli_fetch_assoc($result); 

Answer 2

在您的形式嘗試使用post方法的價值張貼到其他頁面

<form action="details.php" id="searchForm" method="post"> 

嘗試使用請求方法，而不是get方法一樣

$userinput = $_REQUEST['Sample_name'];