Wie Datenbank Wert Textbox angezeigt

Question

// CODE PHP

<?php 
    $conn=mysqli_connect("localhost", "root", "", "dlt"); 
    if(!$conn){ 
    die("Connection Failed: " . mysqli_connect_error()); 
    } 

    if(isset($_POST["searchbutton"])){ 
    $sql="SELECT * FROM professional where LicenceNumber LIKE '".@$_POST['searchbox']."' "; 
    $result=mysqli_query($conn, $sql); 

    if(mysqli_num_rows($result) > 0){ 
    while($row=mysqli_fetch_array($result)){ 

    $licenseno=$row['LicenceNumber']; 
    $fname=$row["FirstName"]; 
    $mname=$row["MiddleName"]; 
    $lname=$row["LastName"]; 
    $adres=$row["Address"]; 
    $bdate=$row["BirthDate"]; 
    $sex=$row["Sex"]; 
    $height=$row["Height"]; 
    $weight=$row["Weight"]; 
    $nationality=$row["Nationality"]; 
    $issue=$row["DateIssued"]; 
    $expiration=$row["ExpirationDate"]; 


    } 
} else { 
    echo "No Record Found"; 
} 
} 
    mysqli_close($conn); 
?> 

// SEARCHBOX UND search

<form action="manage_professional.php" method = "post"> 
<div class="w3-padding-32"> 
<div class="w3-bar w3-card-2 w3-small"> 
<form action = "manage_professional.php" method = "post"> 
    <input class="w3-input w3-border w3-bar-item " type="text" placeholder="Licence No." name="searchbox" maxlength="11" size="65" required> 
    <button name="searchbutton" type="submit" class="w3-bar-item w3-white 
    w3-button w3-border">SEARCH</button> 
    </form> 
    </div> 
    </div> 



    // TEXTBOXES // 3 columns 


    <form action = "manage_professional.php" method = "get"> 
    <div class="w3-third w3-margin-bottom"> 
    <div class="w3-container w3-white"> 
    <p><input value='<?php echo $licenseno?>' name="licenseno" id="licenseno" class="w3-input w3-border" type="text" placeholder="License number" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="First name" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="Middle name" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="Last name" required></p> 
    </div> 
</div> 

<div class="w3-third w3-margin-bottom"> 
    <div class="w3-container w3-white"><p><input class="w3-input w3-border" type="text" placeholder="Address" name="address" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="Birthdate" name="bdate" onfocus="(this.type='date')" onblur="(this.type='text')" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="sex" name="sex" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="Height" name="height" required></p> 
    </div> 
</div> 

<div class="w3-third w3-margin-bottom"> 
    <div class="w3-container w3-white"> 
    <p><input class="w3-input w3-border" type="text" placeholder="Weight" name="weight" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="Nationality" name="nationality" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="Date issued" name="issue" onfocus="(this.type='date')" onblur="(this.type='text')" required></p> 
    <p><input class="w3-input w3-border" type="text" placeholder="Expiration date" name="expiration" onfocus="(this.type='date')" onblur="(this.type='text')" required></p> 
    </div> 
</div> 

<button type="submit" class="w3-button w3-black w3-margin-bottom w3-round" name = "addbutton" style="width:20%">EDIT</button> 
<button type="submit" class="w3-button w3-black w3-margin-bottom w3-round" name = "delbutton" style="width:20%">DELETE</button> 
</form> 
</div> 
</form> 

mir bitte helfen): Ich habe nur einen Code value='<?php echo $licenseno?>' für Versuch und Irrtum jedes Mal hinzugefügt das Formular lädt ich bekomme eine Fehlermeldung

Hinweis: Undefinierte Variable: licenseno in C: \ wamp \ www \ manage_professional.php on line 152

Answer 1

standardmäßig, wenn das Formular erstes Mal geladen wird, wird es nicht if (isset ($ _ POST [ "search"])) Zustand und $ geht in licenseno wird nicht definiert Fehler zeigen, wenn in anderem Fall (isset ($ _ POST [ "search"])) Satz $ licenseno auf null diesem Fehler

Answer 2

Anregung zu vermeiden Sie sollten nicht nur search aber searchbox überprüfen, indem isset