我不能让它工作,我真的不知道如何调试它。在URL中,我要添加以下内容:?候选资格=41(&P);selectCandidate=selectCandidate
global $wpdb;
if(isset($_POST[\'selectCandidate\']))
{
global $wpdb;
$selected_candidate = $_POST[\'candidat\']; // Storing Selected Value In Variable
$wpdb->query( $wpdb->prepare("UPDATE pp_candidates
SET candidate_approved = 1 WHERE candidate_ID = \'".$selected_candidate."\' "));
$wpdb->show_errors();
}
if (!isset($_POST[\'candidatebutton\']))
{echo \'<div class="projectcandidatelist"></div>\';
}
else
{
echo \'<div class="projectcandidatelist"><form action="" id="candidat_list">\';
$project_candidat = $wpdb->get_results("SELECT * FROM pp_candidates WHERE project_ID =\'".$_POST[\'candidatebutton\']."\' ");
foreach($project_candidat as $candidat)
{
echo \'<p><input type="radio" name="candidat" value="\'.$candidat->candidate_ID.\'" >
<a href="\'.$candidat->candidate_url.\'">\'.$candidat->candidate_displayname.\'</a></p>\';
}
echo \'<input type="submit" value="selectCandidate" name="selectCandidate">\';
echo \'</form></div>\';
}
我希望(tinyint)“候选人批准”一栏转到1。提前谢谢。
最合适的回答,由SO网友:Vincent 整理而成
我认为应该是:
$selected_candidate = $_POST[\'candidat\'];
=>
$selected_candidate = $_POST[\'selectCandidate\']
顺便说一句,您正在使用;“准备”;以错误的方式。您没有保护自己免受sql注入的影响,请这样使用:
$wpdb->query( $wpdb->prepare("UPDATE pp_candidates SET candidate_approved = 1 WHERE candidate_ID = %d", $selected_candidate));