我使用twitter typeahead和PHP作为后端从
mysql获取数据.但是当我开始在文本框中输入时,我无法看到任何建议.我认为因为PHP输出必须是
JSON编码..
@H_502_16@我该怎么编码输出
输出:
echo '<a href="results.html" class="searchres" onclick="navigate()" style="color:#696969;text-decoration:none;"><img src='.$iurl.' class="icons" /><div class="results" style="color:#696969;text-decoration:none;">'."$fname".'</div><span style="color:#696969;text-decoration:none;" class="author">'.$caption.'</span></a>'."\n";
HTML:
<!DOCTYPE html> <html lang="en"> <head> <Meta charset="UTF-8"> <title>Example of Twitter Typeahead</title> <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script> <script type="text/javascript" src="../js/typeahead.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ $('input.typeahead').typeahead({ name: 'countries',prefetch: 'getvalues.PHP',limit: 10 }); }); </script> <style type="text/css"> .bs-example{ font-family: sans-serif; position: relative; margin: 100px; } .typeahead,.tt-query,.tt-hint { border: 2px solid #CCCCCC; border-radius: 8px; font-size: 24px; height: 30px; line-height: 30px; outline: medium none; padding: 8px 12px; width: 396px; } .typeahead { background-color: #FFFFFF; } .typeahead:focus { border: 2px solid #0097CF; } .tt-query { Box-shadow: 0 1px 1px rgba(0,0.075) inset; } .tt-hint { color: #999999; } .tt-dropdown-menu { background-color: #FFFFFF; border: 1px solid rgba(0,0.2); border-radius: 8px; Box-shadow: 0 5px 10px rgba(0,0.2); margin-top: 12px; padding: 8px 0; width: 422px; } .tt-suggestion { font-size: 24px; line-height: 24px; padding: 3px 20px; } .tt-suggestion.tt-is-under-cursor { background-color: #0097CF; color: #FFFFFF; } .tt-suggestion p { margin: 0; } </style> </head> <body> <div class="bs-example"> <input type="text" class="typeahead tt-query" autocomplete="off" spellcheck="false"> </div> </body> </html>
getvalues.PHP
<?PHP require_once "config.PHP"; $q = strtolower($_GET["q"]); if (!$q) return; $sql = "select file_name,img_url,captions from completer"; $rsd = MysqL_query($sql); while($rs = MysqL_fetch_array($rsd)) { $fname = $rs['file_name']; $iurl = $rs ['img_url']; $caption = $rs ['captions']; echo '<a href="results.html" class="searchres" onclick="navigate()" style="color:#696969;text-decoration:none;"><img src='.$iurl.' class="icons" /><div class="results" style="color:#696969;text-decoration:none;">'."$fname".'</div><span style="color:#696969;text-decoration:none;" class="author">'.$caption.'</span></a>'."\n"; } ?>