从php文件中创建带有两个按钮的表单以选择操作
[英]Make a form with two buttons from a php file to choose an action
问题描述
i have the following html: 
我有以下html：
<body onload="showcontent()"> <!-- onload optional -->
<div id="content"><img src="loading.gif"></div> <!-- leave img out if not onload -->
</body>
I also have the following script: 我也有以下脚本：
function showcontent(){ 函数showcontent（）{
if(window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject('Microsoft.XMLHTTP');
}
xmlhttp.onreadystatechange = function() {
if(xmlhttp.readyState == 1) {
document.getElementById('content').innerHTML = "<img src='loading.gif' />";
}
if(xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById('content').innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open('GET', 'elsevier.php', true);
xmlhttp.setRequestHeader('Content-type','application/x-www-form-urlencoded');
xmlhttp.send(null);
}
In elsevier.php I want when the rows of a table exceed 500 two buttons to be displayed, telling me whether to continue or cancel. 在elsevier.php中，我希望当表的行数超过500时显示两个按钮，告诉我是继续还是取消。 If I put this code in the .php file nothing happens..... (except that two buttons appear). 如果我将此代码放在.php文件中，则什么也不会发生.....（除了出现两个按钮）。
if(mysqli_errno($con)==1062){
@$q55="SELECT COUNT(*) FROM testarticles";
$result55 = mysqli_query($con, $q55);
$row = $result55->fetch_row();
echo '#: ', $row[0].'<br>';
if($row[0]>=500&&$answer==false){
echo '<form action="" method="GET">';
echo "<label>Do you want to continue or cancel?</label>";
echo '<input type="button" id="but1" name="but1" value="Continue">';
echo '<input type="button" id="but2" name="but2" value="Cancel">';
echo '</form>';
//sleep(2);
if(isset($_GET["but1"])){
$answer=true;
break;
}
elseif(isset($_GET["but2"])){
@$q56="DELETE * FROM journal, volume, issue, articles, testarticles WHERE import_time=$unixtimestamp";
$result56 = mysqli_query($con, $q56);
exit;
}
}
I want in this step the execution of the php script to stop and to display me the two buttons to choose from. 我想在这一步中停止php脚本的执行，并向我显示两个按钮供您选择。 If i press continue i want the script to execute from where it was stopped. 如果我按继续，我希望脚本从停止处执行。
Has anybody any idea?? 有人知道吗？ Thank you in advance! 先感谢您！
2 个解决方案
解决方案1
0 2012-09-24 09:45:38
session_start();
if(mysqli_errno($con)==1062){
$where = isset($_SESSION['last_id']) ? " WHERE id > '" . $_SESSION['last_id'] . "'" : "";
if(isset($_SESSION['last_id']))
unset($_SESSION['last_id']);
@$q55="SELECT COUNT(*) FROM testarticles$where ORDER BY id";
$result55 = mysqli_query($con, $q55);
$row = $result55->fetch_row();
echo '#: ', $row[0].'<br>';
if($row[0] >= 500 && $answer == false){
$_SESSION['last_id'] = $row['id'];
echo "<label>Do you want to continue or cancel?</label>";
echo '<input type="button" id="but1" name="but1" value="Continue" onclick="showcontent(true)" />';
echo '<input type="button" id="but2" name="but2" value="Cancel">';
//sleep(2);
if(isset($_GET["but1"])){
$answer=true;
break;
}
elseif(isset($_GET["but2"])){
@$q56="DELETE * FROM journal, volume, issue, articles, testarticles WHERE import_time=$unixtimestamp";
$result56 = mysqli_query($con, $q56);
exit;
}
}
You have to modify also the showcontent function because when you press the continue button the content retrieved should be appended and not replaced 您还必须修改showcontent函数，因为当您按下继续按钮时，检索到的内容应该附加而不是替换。
function showcontent(){
if(window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject('Microsoft.XMLHTTP');
}
xmlhttp.onreadystatechange = function() {
if(xmlhttp.readyState == 1 && !update) {
document.getElementById('content').innerHTML = "<img src='loading.gif' />";
}
if(xmlhttp.readyState == 4 && xmlhttp.status == 200) {
if(update)
alert("this is an update and I should append data, not to replace the div content");
else {
document.getElementById('content').innerHTML = xmlhttp.responseText;
//this is the first call of the function so the content of the div will be replaced
}
}
}
xmlhttp.open('GET', 'elsevier.php', true);
xmlhttp.setRequestHeader('Content-type','application/x-www-form-urlencoded');
xmlhttp.send(null);
}
HTML Code: HTML代码：
<body onload="showcontent(false)"> <!-- onload optional -->
<div id="content"><img src="loading.gif"></div> <!-- leave img out if not onload -->
</body>
Note that this is not a verified code so you should modify it 请注意，这不是经过验证的代码，因此您应该对其进行修改
解决方案2
-1 2014-11-12 23:31:59
use ' echo ' followed by the < html code > for creating a button. 使用' echo '后跟< html code >来创建按钮。 You will be done. 您将完成。
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