dql7588 2015-08-03 12:44
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AJAX实时搜索JSON数组为空

I try to include a live search into my website. This is my code:

index.html:

<script src="typeahead.min.js"></script>
<input type="text" name="typeahead" class="typeahead tt-query" autocomplete="off" spellcheck="false" placeholder="Type your Query">

search.php

<?php
    $key=$_GET['key'];
    $array = array();
    $con=mysql_connect("localhost","root","");
    $db=mysql_select_db("gs_dev",$con);
    $query=mysql_query("select * from dmdkh_posts where post_title LIKE '%{$key}%'");
    $query2=mysql_query("select * from dmdkh_posts where post_title LIKE '%in%'");

    //THIS OUTPUT WORKS! I get all titles.
    //echo $query2;
    /*while ($row = mysql_fetch_array($query2, MYSQL_ASSOC)) {
    printf("Title: %s ", $row["post_title"]);
    }*/ 

    while($row=mysql_fetch_array($query, MYSQL_ASSOC))
    {
      $array[] = $row['post_title'];
    }
    //THIS OUTPUT IS EMPTY! OUTPUT: []
    echo json_encode($array);
?>

I've two query.. (query and query2) the ouput of query2 works, I get the titles. The output of query is "[]"

In firebug I can see the parameters which will transfer from index.html into search.php.

I hope someone can guard me.

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1条回答 默认 最新

  • dougong1031 2015-08-03 13:14
    关注

    Change your first query with $query=mysql_query("select * from dmdkh_posts where post_title LIKE '%$key%'");

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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