php搜索查询只输出两个结果...而不是整行

时间:2015-03-12 10:35:52

标签: php mysql

您好我在index.html中有一个搜索查询..当用户按下搜索时...它会转到searchjobs.php。

我希望输出整行。这是 JOB_ID 职称 职位描述 工作地点 job_category

目前它只输出其中两个字段,即job_title和job _description。如何让它输出整行?例如,如果此人搜索“利兹”,我希望它能够在利兹中输出关于作业的整行。

我认为问题出在回声区域。

如果我在php脚本中输入更多变量......它将无法运行。

<?php
    mysql_connect("*****", "root", "*****") or die("Error connecting to database: ".mysql_error());
    /*

        if connection fails it will stop loading the page and display an error
    */

    mysql_select_db("jobslist") or die(mysql_error());
    /* tutorial_search is the name of database we've created */


    $query = $_GET['query'];
    // gets value sent over search form

    $min_length = 4;
    // you can set minimum length of the query if you want

    if(strlen($query) >= $min_length){ // if query length is more or equal minimum length then

        $query = htmlspecialchars($query);
        // changes characters used in html to their equivalents, for example: < to &gt;

        $query = mysql_real_escape_string($query);
        // makes sure nobody uses SQL injection

        $raw_results = mysql_query("SELECT * FROM jobs_list
            WHERE (`job_title` LIKE '%".$query."%') OR (`job_location` LIKE '%".$query."%')") or die(mysql_error());

        // * means that it selects all fields, you can also write: `id`, `title`, `text`
        // jobs_list is the name of our table

        // '%$query%' is what we're looking for, % means anything, for example if $query is Hello
        // it will match "hello", "Hello man", "gogohello", if you want exact match use `title`='$query'
        // or if you want to match just full word so "gogohello" is out use '% $query %' ...OR ... '$query %' ... OR ... '% $query'

        if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following

            while($results = mysql_fetch_array($raw_results)){
            // $results = mysql_fetch_array($raw_results) puts data from database into array, while it's valid it does the loop

                echo "<p><h3>".$results['job_id']."</h3>"
                              .$results['job_title'].
                                .$results['job_description'].
                                .$results['job_location'].
                                .$results['job_category'].



                           "</p>";
                // posts results gotten from database you can also show id ($results['id'])
            }

        }
        else{ // if there is no matching rows do following
            echo "No results";
        }

    }
    else{ // if query length is less than minimum
        echo "Minimum length is ".$min_length;
    }
?>

1 个答案:

答案 0 :(得分:2)

您在变量的两侧使用了连接点,删除了多余的点:

echo '<p><h3>'.$results['job_id'].'</h3>'
     . $results['job_title']
     . $results['job_description']
     . $results['job_location']
     . $results['job_category']
     . '</p>';

我假设你正在测试和尝试,否则你应该正确地构建你的工作描述,即

$tplJob= '<div class="job_entry">
  <h3>ID: %s</h3>
  <div class="title">%s</div>
  <div class="description">%s</div>
  <div class="location">Location: %s</div>
  <div class="category">Category: %s</div>
</div>';

// echo formatted output based on above template variable
echo sprintf($tplJob, $results['job_id'], $results['job_title'],  $results['job_description'], $results['job_location'], $results['job_category']);