wie man Daten von der Datenbank im Textkasten in php zeigt

Question

ich bin neu in php Programmierung ich habe Problem in diesem Code, ich will Daten im Textfeld zeigen, wenn ich Knopf wähle, der im Gitter ist, aber ich bin darauf fest da die Daten nicht in Textfeld und meine Select-Taste arbeitet auch nicht gezeigt, was für das ich tun Freunde, ist dies mein Code

VORNAME ZWEITER VORNAME NAME

**--//php code--//** 

$qrydatabind='SELECT ecode, first_name, middle_name, last_name, father_name, mother_name, 
      number_of_dependents, dob, gender, identification_mark, marital_status, spouse_name, mobile_number, 
      email_id, adhar_id, pan_number, passport_number, tin_number, dl_number FROM USER_MASTER ORDER BY user_id DESC 
      LIMIT 1'; 
        $results= mysql_query($qrydatabind) or die(mysql_error()); 

         while($row = mysql_fetch_array($results)) { 
         echo 
      " 
       <div class='table-responsive'> 
<table border='1' style= 'background-color: #84ed86; color: #761a9b; ' > 
     <thead> 
       <tr> 
      <th></th> 
       <th>Entity Code</th> 
       <th>User Id</th> <th>User Name</th> <th>Father Name</th> <th>Mother Name</th> <th>No.Of Dependents</th> 
       <th>D.O.B</th> <th>GENDER</th> <th>Id Mark</th> <th>MARITAL STATUS</th> <th>SPOUSE NAME</th> 
       <th>Mob. Number</th> <th>E-Id</th> <th>ADHAR-ID</th> <th>PAN-No.</th> <th>PASSPORT-No.</th> 
       <th>TIN-NO.</th> <th>DL-No.</th> 
      </tr> 
       </thead> 

         <tr >   

         <td><button type='submit' class='btn btn-primary' name='select_button'>Select</button> </td> 
         <td>{$row['ecode']}</td> <td> echo $row[0];</td> 
         <td>{$row['first_name']} {$row['middle_name']} {$row['last_name']}</td>    
         <td>{$row['father_name']}</td> <td>{$row['mother_name']}</td> 
         <td>{$row['number_of_dependents']}</td> <td>{$row['dob']}</td> 
         <td>{$row['gender']}</td> <td>{$row['identification_mark']}</td> 
         <td>{$row['marital_status']}</td> <td>{$row['spouse_name']}</td> 
         <td>{$row['mobile_number']}</td> <td>{$row['email_id']}</td> 
         <td>{$row['adhar_id']}</td> <td>{$row['pan_number']}</td> 
         <td>{$row['passport_number']}</td> <td>{$row['tin_number']}</td> 
         <td>{$row['dl_number']}</td> 

      </tr> </table> 
       </div>"; 
     }} 

     if(isset($_POST['select_button'])) 
      { 
      $qrydatabind1='SELECT ecode, first_name, middle_name, last_name, father_name, mother_name, 
      number_of_dependents, dob, gender, identification_mark, marital_status, spouse_name, mobile_number, 
      email_id, adhar_id, pan_number, passport_number, tin_number, dl_number FROM USER_MASTER'; 
        $results1= mysql_query($qrydatabind1) or die(mysql_error()); 

         while($row = mysql_fetch_array($results1)) { 
          echo' 
         <tr >     
         <td> </td> 

         <td><input type="text" value="{$row["first_name"]}" ></td> 

      </tr>'; 

      } 
      } 

Answer 1

Es ist sehr einfach. wie Sie Daten eingeben.

Ihr Code

<td> echo $row[0];</td> 

Textfeld Code:

<input type="text" value="<?php echo $row[0];?>"> 

Answer 2

PHP-Variablen werden nur in Strings ausgewertet werden, die in doppelte Anführungszeichen zitiert werden "

Sie entweder einen String concatinate sollte:

echo '<td><input type="text" value="' . $row["first_name"] . '" ></td>'; 

Oder

, verwenden <?php und ?>

while($row = mysql_fetch_array($results1)) { 

    ?><td><input type="text" value="<?php echo $row["first_name"]; ?>" ></td><?php 

} 

Answer 3

You can also use this as 
 
echo'<tr><td><input type="text" value="'.$row["first_name"].'" ></td></tr>';