sufibd2010 Posted March 26, 2012 Share Posted March 26, 2012 if ($showigiteidimage) { $igitwid_divlnk = "onclick=location.href='" . get_permalink($igitwid_result->ID) . "'; style=cursor:pointer;"; $igitwid_output .= '<li id="igit_wid_rpwt_li" style="height:' . $igitwid_height . 'px; padding-bottom: 10px;" ' . $igitwid_divlnk . '>'; $igitwid_output .= '<div id="igit_wid_rpwt_main_image" style="float:left;"> <a href="' . get_permalink($igitwid_result->ID) . '" target="_top"><img src="<?php $values = get_post_custom_values("tj_video_img_url"); echo $values[0]; ?>" id="igit_rpwt_thumb" ></a></div>'; the prb is image src cant read the php code , what to do ? please helpppppppppp :( :( :( :'( :'( Quote Link to comment https://forums.phpfreaks.com/topic/259754-cant-detect-code/ Share on other sites More sharing options...
gristoi Posted March 27, 2012 Share Posted March 27, 2012 your trying to echo php within a string you are already building in php. try: <?php $imgsrc = get_post_custom_values("tj_video_img_url"); $igitwid_output .= '<div id="igit_wid_rpwt_main_image" style="float:left;"> <a href="' . get_permalink($igitwid_result->ID) . '" target="_top"><img src="'. $imgsrc[0] .'" id="igit_rpwt_thumb" ></a></div>'; as the last two lines of your code Quote Link to comment https://forums.phpfreaks.com/topic/259754-cant-detect-code/#findComment-1331566 Share on other sites More sharing options...
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