<script>
<?php
mysql_connect('127.0.0.1', 'root', '') or die(mysql_error());
mysql_select_db("Project_Part1")or die("cannot select DB");
$sqlActivity = "SELECT aname FROM Activity";
$resultActivity=mysql_query($sqlActivity);
$aname = array();
while ($row = mysql_fetch_array($resultActivity)) {
$aname[] = $row;
}
?>
$(function() {
var availableTags =[ "<?php echo implode('","',$aname);?>" ];
$( "#tags" ).autocomplete({
source: availableTags
});
});
</script>
<div class="ui-widget">
<label for="tags">Tags: </label>
<input id="tags" />
</div>不确定为什么var availableTags =[ "<?php echo implode('","',$aname);?>" ];不能自动完成。当我使用var availableTags = [ "Rock climbing","Fishing","Kayaking","Underwater bungee jumping" ];时,它是好的。
我是PHP新手。有人能在这方面帮我吗?
发布于 2014-05-05 03:41:40
尝试:
<?php
$DBi = mysqli_connect($hostname, $user, $password, $database);
$aname = array();
$sqlActivity = "SELECT `aname` FROM `Activity`";
$resultActivity = mysqli_query($DBi, $sqlActivity); //Dump mysql_query for
//mysqli_query and don't
//forget the connection bit
while ($row = mysqli_fetch_array($resultActivity)) {
$aname[] = $row;
};
?>
<script>
$(function() {
var availableTags =[ "<?php echo implode('","',$aname);?>" ];
$("#tags").autocomplete({
source: availableTags
});
});
</script>
<div class="ui-widget">
<label for="tags">Tags: </label>
<input id="tags" />
</div>看起来你在你的<script>标签后面缺少<?php。我将PHP SQL位移出了您的<script>标记(这应该无关紧要,但确实有助于让事情在我的脑海中更容易理解)
在此过程中,请停止使用mysql_并切换到mysqli_函数,因为前者已弃用
https://stackoverflow.com/questions/23460678
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