Wie zu definieren, welches Bild in HTML-Seite angeklickt und speichern Sie den Bildnamen in MySQL mit PHP

mein Projekt ist Web-Bild Re-Ranking. Ich versuche zu versuchen, welches Bild in der HTML-Seite ausgewählt wurde, um den Rang des Bildes in Mysql zu aktualisieren. Mein Code istWie zu definieren, welches Bild in HTML-Seite angeklickt und speichern Sie den Bildnamen in MySQL mit PHP

index.php

<?php 


include './search_image.php'; 

$obj_image = new Image(); 

if(@$_POST['Submit']) 
{ 
$obj_image->image_name=str_replace("'", "''", $_POST['txt_image_name']); 

$obj_image->Insert_into_image(); 

$data_image=$obj_image->get_all_image_list(); 
$row=mysql_num_rows($data_image); 
} 

?> 



    <!DOCTYPE HTML PUBLIC"-//W3C//DTD HTML 4.0 Transitional//EN"> 
    <html> 
    <head> 
    <title>Home</title> 
    </head> 
    <body bgcolor=cyan> 
    <script> 
    window.location.hash="no-back-button"; 
    window.location.hash="Again-No-back-button";//again because google chrome   don't insert first hash into history 
    window.onhashchange=function(){window.location.hash="no-back-button";} 
    </script> 
    <center><h1>WEB IMAGE RERANKING</H1></CENTER> 
    <center> 
    <table border=0 cellspacing=10> 
    <th><a href="index.php" target="right">HOME</th> 
    <th><a href="admin.php" target"=right">ADMIN</th> 
    <form method="post" enctype="multipart/form-data"> 
    <tr> 
    <th width="50%">Enter Image name for search</th> 
    <td width="50%"><input type="text" name="txt_image_name"></input></td> 
    </tr> 
    <td width="50%"><input type="submit" name="Submit" value="Submit"></input>  </td> 
    </tr> 
    </table> 
    </form> 
    </CENTER> 




    <?php 
    if($row!=0) 
    { 
    ?> 
    <center> 
    <form method="post" enctype="multipart/form-data"> 

    <table width="30%" border="0"> 
    <?php 

    while($data= mysql_fetch_assoc($data_image)) 
    { 
    ?> 
    <tr> 

    <a href=# onclick="loadImage(String $data['image'])"> <img src="images/<?php echo $data['image']; ?>" width="400px" height="200px" ></a> 
    </tr> 
    <?php 
    } 
    ?> 
    </table> 
    </form> 
    </center> 
    <?php 
    } 

    ?> 
    <?php 
    echo" <script language='javascript' type='text/javascript'> 
function loadImage(String a) { 

     alert('1 %s',a); 
    } 
    </script>" 
    ?> 
    </body> 
    </html>

search_image.php

<?php 
     include 'db_connection.php' ; 

     class Image{ 

    var 
    $image_id, 
     $image_name, 
     $image; 

     function Insert_into_image(){ 

     } 

     function get_all_image_list(){ 
     $query = "select *from stor_image where img_name='$this->image_name'"; 
     $result = mysql_query($query); 
     return $result; 
     } 

     } 
     ?>

wenn Klick auf bestimmtes Bild den Rang des Bildes aktualisieren. Also brauche ich Hilfe, um Bildname oder src zu finden, um die Daten in mysql Datenbank zu aktualisieren. Danke

Quelle

2016-07-31 Sukhesh Chukkapalli

HTML: 
<a href="javascript:void(0);" class="gallery" id="<?php echo $data['image']; ?>" onClick="reply_click(this.id)"><img src="images/<?php echo $data['image']; ?>" width="400px" height="200px" ></a> 
<script> 
function reply_click(clicked_id) 
{ 
    $imagename=clicked_id; 
    alert($imagename); 
} 
</script>

Quelle

2016-07-31 17:56:11

Try this:

HTML: 
<a href="javascript:void(0);" class="gallery" id="<?php echo $data['image']; ?>"><img src="images/<?php echo $data['image']; ?>" width="400px" height="200px" ></a> 

JS: 
$('.gallery').click(function(){ 
var image_name = $(this).attr('name'); 

// make an ajax call with this image name and do what ever you want to do 
});

Quelle

2016-07-31 15:04:04

Dank sehr viel. –

Wie zu definieren, welches Bild in HTML-Seite angeklickt und speichern Sie den Bildnamen in MySQL mit PHP

Antwort

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