Question

所以我不是那种在编程方面经验丰富的人，我正在开发一些php。

我的查询（不包括我破坏的if-else语句＆gt; _＆gt;），但是当我提交1个查询（例如query2）时，它可以打印结果，以及另一个查询7的结果。我怎么能阻止它？

此外，如果有人知道我在第一个查询和查询6的if-else语句中失败了，我会感激一些见解（他们都使用html提交按钮）

谢谢！

这是我的问题php代码：

$lastName = $_POST['lastName'];

if ($_Post['lastName'] = "") {
    $query = "SELECT c.*, s.speaker_year FROM Contact c, Speakers s WHERE s.Contact_con_id = c.con_id ";
} Else {
    $query = "SELECT c.*, s.speaker_year FROM Contact c, Speakers s WHERE s.Contact_con_id = c.con_id 
AND con_lname =  ";
}
$query = $query . "'" . $lastName . "' ORDER BY con_lname;";

$rgroups = $_POST['rgroups'];

if ($_Post['rgroups'] = "") {
$query6 = "SELECT r.rev_groups_id, c.con_fname, c.con_lname, con_phone, rev_groups_pass, count(p.proposal_id) 
FROM Review_Groups r JOIN Proposal p on r.rev_groups_id = p.Review_Groups_rev_groups_id
JOIN Presents px on px.Proposal_proposal_id = p.proposal_id
JOIN Contact c on px.Speakers_Contact_con_id = c.con_id
JOIN Reviewer rw on rw.Review_Groups_rev_groups_id = r.rev_groups_id
WHERE rw.reviewer_type = 'local'
AND r.rev_groups_id = ";
$query6 = $query6 . "'" . $rgroups . "' Group BY r.rev_groups_id;";} 
Else {
$query6 = "SELECT r.rev_groups_id, c.con_fname, c.con_lname, con_phone, rev_groups_pass, count(p.proposal_id) 
FROM Review_Groups r JOIN Proposal p on r.rev_groups_id = p.Review_Groups_rev_groups_id
JOIN Presents px on px.Proposal_proposal_id = p.proposal_id
JOIN Contact c on px.Speakers_Contact_con_id = c.con_id
JOIN Reviewer rw on rw.Review_Groups_rev_groups_id = r.rev_groups_id
WHERE rw.reviewer_type = 'local'
AND r.rev_groups_id = ";

$query6 = $query6 . "'" . $rgroups . "' ";}

$check = $_POST['check'];

$query7 = "Select c.con_fname, c.con_lname, s.Contact_con_id,
IF(s.Contact_con_id IS NULL, 'NO', 'YES')
From Contact c Left Join (Select Contact_con_id FROM Speakers
WHERE speaker_year = '". $check . "') As s
ON c.con_id = s.Contact_con_id";
$query7 = $query7 . " ORDER BY c.con_fname;";
(this is the code that prints on every result)

$average = $_POST['average'];

$query5 = "SELECT c.con_fname, r.Reviewer_Contact_con_id, question_id, AVG( DISTINCT question_score)
FROM Contact c, Individual_Review r
WHERE r.Reviewer_Contact_con_id = c.con_id
AND con_fname = ";
$query5 = $query5 . "'" . $average . "' GROUP BY r.Proposal_proposal_id;";
(example of working code. you can put in George next to con_fname to get a result)

Answer 1

// 1.  Format your code with indents, etc.
// 2. Comment your code
// 3. Don't pass $_POST data straight to your sql.
// 4. Variables are case sensitive, including POST

$lastName = $_POST['lastName'];

if ($lastName = "") {
    $query = "SELECT c.*, s.speaker_year FROM Contact c, Speakers s WHERE           
                s.Contact_con_id = c.con_id ";
}else{
    $query = "SELECT c.*, s.speaker_year FROM Contact c, Speakers s WHERE 
                        s.Contact_con_id = c.con_id 
                        AND con_lname =  ";
}
$query = $query . "'" . $lastName . "' ORDER BY con_lname;";
// if you did the first if, then this broke.
// Use:
// echo $query; 
// to see what you have so far.  
    $query = "SELECT c.*, s.speaker_year FROM Contact c, Speakers s WHERE 
                        s.Contact_con_id = c.con_id 
                        AND con_lname =  '".$lastName."' ORDER BY con_lname";

$rgroups = $_POST['rgroups'];

// you can go like $query .= 
// you don't have to do $query = $query;
// so all of this could be:  
$query6 = "SELECT r.rev_groups_id, c.con_fname, c.con_lname, con_phone, rev_groups_pass, count(p.proposal_id) 
FROM Review_Groups r JOIN Proposal p on r.rev_groups_id = p.Review_Groups_rev_groups_id
JOIN Presents px on px.Proposal_proposal_id = p.proposal_id
JOIN Contact c on px.Speakers_Contact_con_id = c.con_id
JOIN Reviewer rw on rw.Review_Groups_rev_groups_id = r.rev_groups_id
WHERE rw.reviewer_type = 'local' ";

if ($_Post['rgroups'] = "") {
    $query6 .= " AND r.rev_groups_id = '" . $rgroups . "' Group BY r.rev_groups_id;";
}else{
    $query6 = "SELECT r.rev_groups_id, c.con_fname, c.con_lname, con_phone, rev_groups_pass, count(p.proposal_id) 
                FROM Review_Groups r JOIN Proposal p on r.rev_groups_id = p.Review_Groups_rev_groups_id
                JOIN Presents px on px.Proposal_proposal_id = p.proposal_id
                JOIN Contact c on px.Speakers_Contact_con_id = c.con_id
                JOIN Reviewer rw on rw.Review_Groups_rev_groups_id = r.rev_groups_id
                WHERE rw.reviewer_type = 'local'
                AND r.rev_groups_id = '" . $rgroups . "' ";
}

$check = $_POST['check'];

Answer 2

您可以在某些if条件中添加$ query7以避免

Answer 3

注意：我只处理您的PHP结构。我根本没有看过你的SQL语法。但是我给了你工具来看看SQL是否正在返回你认为应该返回的内容。

<?PHP

// here are some functions for ya
function sqlarr($sql, $numass=MYSQL_BOTH) {
    // MYSQL_NUM  MYSQL_ASSOC  MYSQL_BOTH
    $got = array();
    $result=mysql_query($sql) or die("$sql: " . mysql_error());                             

    if(mysql_num_rows($result) == 0)
        return $got;
    mysql_data_seek($result, 0);
    while ($row = mysql_fetch_array($result, $numass)) {
        array_push($got, $row);
    }
    return $got;
} 

// Sql fetch assoc
function sqlassoc($sql){
    $query = mysql_query($sql) or die("$sql:". mysql_error());
    $row = mysql_fetch_assoc($query);
    return $row;
}

function sqlrow($sql){
    $query = mysql_query($sql) or die("$sql:". mysql_error());
    $row = mysql_fetch_row($query);
    return $row;
}

function sqlquery($sql){
    $query = mysql_query($sql) or die("$sql:". mysql_error());
    return $row;
}

function printr( array $array, $label = '' ){
    echo '<pre>'.$label;
    print_r( $array );
    echo '</pre>';  
}

// This isn't the best, but it's better than nothing
// use PDO when you get more advanced
function makeSomewhatSafe($str){
    return htmlspecialchars(stripslashes(strip_tags($str, '<p>')), ENT_QUOTES);
}

// good practice:  initiate any variables you use at the beginning
// we're going to go ahead and strip them here too to try to avoid sql injection
$rgroups = makeSomewhatSafe($_POST['rgroups'] );
$lastName = makeSomewhatSafe( $_POST['lastName'] );
$query = NULL;
$speakerContactResulst = array();
$check = makeSomewhatSafe( $_POST['check'] );
$average = makeSomewhatSafe($_POST['average']);

// if($_Post['lastName'] = "") {
// we're going to see if it has a value
// another way to do this if your empty isn't working is to do 
// if( strlen( $lastName ) > 0 ){
if( empty( $lastName ) ){
    $query = "SELECT c.*, s.speaker_year FROM Contact c, Speakers s WHERE s.Contact_con_id = c.con_id ";
}else{
    $query = "SELECT c.*, s.speaker_year FROM Contact c, Speakers s WHERE s.Contact_con_id = c.con_id 
AND con_lname =  ";
}
$query .= "'" . $lastName . "' ORDER BY con_lname";

     echo 'This query states: '.$query.' <br /><br />';
$speakerContactResulst = sqlarr( $query );
printr( $speakerContactResulst, 'speakerContactResulst ');

if ( ! empty( $rgroups ) ){
    $query = "SELECT r.rev_groups_id, c.con_fname, c.con_lname, con_phone, rev_groups_pass, count(p.proposal_id) 
    FROM Review_Groups r JOIN Proposal p on r.rev_groups_id = p.Review_Groups_rev_groups_id
    JOIN Presents px on px.Proposal_proposal_id = p.proposal_id
    JOIN Contact c on px.Speakers_Contact_con_id = c.con_id
    JOIN Reviewer rw on rw.Review_Groups_rev_groups_id = r.rev_groups_id
    WHERE rw.reviewer_type = 'local'
    AND r.rev_groups_id = '" . $rgroups . "' Group BY r.rev_groups_id;";
}else{
// I dont know if you matters, but keep your else's more compact.  Don't do like you had with the else on a new line
// str'; }
// else {
    $query = "SELECT r.rev_groups_id, c.con_fname, c.con_lname, con_phone, rev_groups_pass, count(p.proposal_id) 
    FROM Review_Groups r JOIN Proposal p on r.rev_groups_id = p.Review_Groups_rev_groups_id
    JOIN Presents px on px.Proposal_proposal_id = p.proposal_id
    JOIN Contact c on px.Speakers_Contact_con_id = c.con_id
    JOIN Reviewer rw on rw.Review_Groups_rev_groups_id = r.rev_groups_id
    WHERE rw.reviewer_type = 'local'
    AND r.rev_groups_id = '" . $rgroups . "' ";
}

$groupResults = sqlarr( $query );
printr( $groupResults, 'groupResults' );

$query = "Select c.con_fname, c.con_lname, s.Contact_con_id,
IF(s.Contact_con_id IS NULL, 'NO', 'YES')
From Contact c Left Join (Select Contact_con_id FROM Speakers
WHERE speaker_year = '". $check . "') As s
ON c.con_id = s.Contact_con_id  ORDER BY c.con_fname;";

$checkResults = sqlarr( $query );

$query = "SELECT c.con_fname, r.Reviewer_Contact_con_id, question_id, AVG( DISTINCT question_score)
FROM Contact c, Individual_Review r
WHERE r.Reviewer_Contact_con_id = c.con_id
AND con_fname = '" . $average . "' GROUP BY r.Proposal_proposal_id;";

$averageResults = sqlarr( $query );
?>

PHP打印2个查询集，不需要

3 个答案: