url de imagen en Spintax
Publicado por christian (1 intervención) el 07/01/2016 18:37:28
Buenos dias, estoy un poco liado tratando de que mi web en la parte donde te da los link de las imagenes subidas, te los link de las imagenes en formato Spintax? {URL1.jpg|URL2.jpg|URL3.jpg}
Alguien sabe como puedo hacer??
Alguien sabe como puedo hacer??
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<div id="uploadedimage">
<a target='_blank' href="<?php echo "{$site_url}/img-{$view_id}.html"; ?>"><img border="0" src="<?php echo $site_url . "/" . $resizeuploadpatch; ?>" alt="uploaded_image" /></a>
</div>
<div id="uploadcodes">
<label>Link:</label><br />
<input type='text' onclick="this.select();" value="<?php echo "{$site_url}/{$dir}/{$nameimage}"; ?>">
<br /> <br />
<label>HTML:</label><br />
<input type='text' onclick="this.select();" value="<?php echo "<a href='{$site_url}/img-{$view_id}.html'><img src='{$site_url}/{$dir}/{$nameimage}' alt='image' /></a> "; ?>">
<br /> <br />
<label>Link:</label><br />
<input type='text' onclick="this.select();" value="<?php echo "{$site_url}/img-{$view_id}.html "; ?>">
<?php
if(DIRECT_LINK_SHOW == 1) {
echo " <br /> <br />
<label>Direct Link to image:</label><br />
<input type='text' onclick='this.select();' value='{$site_url}/{$dir}/{$nameimage}'>
"; }?>
</div>
<?php
global $BBCode_global;
global $HTMLCode_global;
global $DirectLink_global;
global $DirectLinkToImg_global;
$BBCode_global[] = "[{$site_url}/{$dir}/{$nameimage}]";
$HTMLCode_global[] = "<a href='{$site_url}/img-{$view_id}.html'><img src='{$site_url}/{$resizeuploadpatch}' alt='image' /></a>";
$DirectLink_global[] = "{$site_url}/img-{$view_id}.html";
$DirectLinkToImg_global[] = "{$site_url}/{$dir}/{$nameimage}";
echo "<div style='display:none;' class='ajax_BBCode'>[URL={$site_url}/img-{$view_id}.html][/URL]</div>"; echo "<div style='display:none;' class='ajax_HTMLCode'><a href='{$site_url}/img-{$view_id}.html'><img src='{$site_url}/{$resizeuploadpatch}' alt='image' /></a></div>"; echo "<div style='display:none;' class='ajax_DirectLink'>{$site_url}/img-{$view_id}.html</div>"; echo "<div style='display:none;' class='ajax_DirectLinkToImg'>{$site_url}/{$dir}/{$nameimage}</div>";} else {
echo "<p class='error'>Sorry, there was a problem uploading your file.</p>"; }Valora esta pregunta
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