如何解决IF和ELSE都不会触发php脚本流!为什么?
伙计,
在条件之前,我并没有杀死脚本流
die();
因此,两个触发之一应触发: 如果 否则
但是他们没有。
if(!$result_2)
{
echo __LINE__; echo "<br>"; //LINE 92: THIS LINE IN 'IF' DOES NOT TRIGGER. NEITHER THE LINE IN THE 'ELSE'. WHY ?
die("Fetching Error");
}
while($row = mysqli_fetch_array($result_2,MYSQLI_ASSOC))
{
echo __LINE__; echo "<br>";//LINE 98: THIS LINE IS LIKE AN 'ELSE'. IT DOES NOT TRIGGER. NEITHER THE LINE IN THE 'IF'. WHY ?
第92行(IF)应该触发,或者第98行(类似于ELSE)触发。
请注意代码中的注释,以了解脚本无故终止流的位置。
<?php
$query_2 = "SELECT id,page_url,link_anchor_text,page_description,keyphrases,keywords FROM links WHERE keywords = ? ORDER by id LIMIT $offset,$limit";
$stmt_2 = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt_2,$query_2))
{
echo __LINE__; echo "<br>";//LINE 84: THIS LINE GETS TRIGGERED AND ECHOED IN TEST. LAST LINE THAT GETS ECHOED.
mysqli_stmt_bind_param($stmt_2,'s',$keywords);
mysqli_stmt_execute($stmt_2);
$result_2 = mysqli_stmt_get_result($stmt_2);
if(!$result_2)
{
echo __LINE__; echo "<br>"; //LINE 92: THIS LINE IN 'ELSE' DOES NOT TRIGGER. NEITHER THE LINE IN THE 'IF'. WHY ?
die("Fetching Error");
}
什么是杀死脚本流?
上下文:
<?php
echo __LINE__; echo "<br>"; //LINE 16: THIS LINE GETS TRIGGERED AND ECHOED IN TEST
if(!ISSET($_GET['keywords']))
{
echo __LINE__; echo "<br>";
die("Type your keywords");
}
$keyword = $_GET['keywords'];
//Check if the PAGE NUMBER is specified or not and if it's a numer or not. If not,return the default: 1.
$page = ISSET($_GET['page']) && is_numeric($_GET['page']) ? $_GET['page'] : 1;
//Check if the PAGE RESULT LIMIT is specified or not and if it's a numer or not. If not,return the default: 1.
$limit = ISSET($_GET['limit']) && is_numeric($_GET['limit']) ? $_GET['limit'] : 1;
$query_1 = "SELECT COUNT(id) FROM links WHERE keywords = ?";
$stmt_1 = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt_1,$query_1))
{
echo __LINE__; echo "<br>";//LINE 57: THIS LINE GETS TRIGGERED AND ECHOED IN TEST. BUT WHY SCRIPT FLOW DOES NOT GO BEYOND THIS LINE ?
mysqli_stmt_bind_param($stmt_1,$keywords);
mysqli_stmt_execute($stmt_1);
$result_1 = mysqli_stmt_bind_result($stmt_1,$row_count);
mysqli_stmt_fetch($stmt_1);
}
else
{
echo __LINE__; echo "<br>"; //LINE 67
printf("Error: %s.\n",mysqli_stmt_error($stmt_1));
printf("Error: %d.\n",mysqli_stmt_errno($stmt_1));
die("A. Prepare failed!");
}
mysqli_stmt_close($stmt_1);
//$total_pages = ceil($result_1/$limit); //SHOULD I KEEP THIS LINE OR THE ONE BELOW THIS ONE ?
$total_pages = ceil($row_count/$limit); //SHOULD I KEEP THIS LINE OR THE ONE ABOVE THIS ONE ?
$offset = (($page * $limit) - $limit);
$query_2 = "SELECT id,$keywords);
mysqli_stmt_execute($stmt_2);
$result_2 = mysqli_stmt_get_result($stmt_2);
if(!$result_2)
{
echo __LINE__; echo "<br>"; //LINE 92: THIS LINE IN 'IF' DOES NOT TRIGGER. NEITHER THE LINE IN THE 'ELSE'. WHY ?
die("Fetching Error");
}
while($row = mysqli_fetch_array($result_2,MYSQLI_ASSOC))
{
echo __LINE__; echo "<br>";//LINE 98: THIS LINE IS LIKE AN 'ELSE'. IT DOES NOT TRIGGER. NEITHER THE LINE IN THE 'IF'. WHY ?
echo "LIMIT: $limit<br>";
echo "ROW COUNT: $row_count<br><br>";
echo "TOTAL PAGES: $total_pages<br><br>";
//Retrieve Values.
$id = $row["id"];
$page_url = $row["page_url"];
$link_anchor_text = $row["link_anchor_text"];
$page_description = $row["page_description"];
$keyphrases= $row["keyphrases"];
$keywords = $row["keywords"];
echo "Id: $id<br>";
echo "Page Url: $page_url<br>";
echo "Link Anchor Text: $link_anchor_text<br>";
echo "Page Description: $page_description<br>";
echo "Keyphrases: $keyphrases<br>";
echo "Keywords: $keywords<br>";
echo "<br>";
echo "<br>";
}
}
else
{
echo __LINE__; echo "<br>";//Line 124
printf("Error: %s.\n",mysqli_stmt_error($stmt_2));
printf("Error: %d.\n",mysqli_stmt_errno($stmt_2));
die("B. Prepare failed!");
}
此外,在这2个中,哪个是正确的?
$total_pages = ceil($result_1/$limit); //SHOULD I KEEP THIS LINE OR THE ONE BELOW THIS ONE ?
$total_pages = ceil($row_count/$limit); //SHOULD I KEEP THIS LINE OR THE ONE ABOVE THIS ONE ?
更新 我的代码现在看起来像这样。 在CAPITALS的代码注释中清楚地提到了我的问题。
当前问题:
问题1: 分页部分仅显示从PAGE 1到PAGE 1的链接。无法显示到PAGE 2的链接。配置为每页显示1行的代码。由于有2个匹配的行,因此自然会有2页显示所有结果。为什么分页部分无法显示从PAGE 1到PAGE 2的链接?
error_reporting.php
<?php
ini_set('display_errors','1');
ini_set('display_startup_errors','1');
ini_set('error_reporting',E_ALL);
?>
<?php
require 'conn.php';
require 'error_reporting.php';
?>
<!DOCTYPE HTML">
<html>
<head>
<meta name="viewport" content="width-device=width,initial-scale=1">
</head>
<body>
<form method='GET' action="<?php echo $_SERVER['PHP_SELF'];?>?keywords=$keywords&limit=$limit&page=1">
<label for="keywords">Keywords:*</label>
<input type="text" name="keywords" id="keywords" placeholder="Input Keywords" required>
<br>
<label for="limit">Results per Page</label>
<select name="limit" id="limit">
<option value="1">1</option>
<option value="10">10</option>
<option value="25">25</option>
<option value="50">50</option>
<option value="100">100</option>
</select>
<br>
<button name="search" id="search" value=" ">Search</button><br>
<button type="submit" name="search" id="search" value="search">Search</button>
<br>
<input type="reset">
<br>
</form>
<?php
echo __LINE__; echo "<br>"; //LINE 16: THIS LINE GETS TRIGGERED AND ECHOED IN TEST
if(!ISSET($_GET['keywords']))
{
echo __LINE__; echo "<br>";
die("Type your keywords");
}
$keywords = $_GET['keywords'];
//Check if the PAGE NUMBER is specified or not and if it's a numer or not. If not,$query_1))
{
echo __LINE__; echo "<br>";//LINE 57: THIS LINE GETS TRIGGERED AND ECHOED IN TEST.
mysqli_stmt_bind_param($stmt_1,$row_count);
mysqli_stmt_fetch($stmt_1);
echo "ROW COUNT: $row_count<br><br>";
}
else
{
echo __LINE__; echo "<br>"; //LINE 67
printf("Error: %s.\n",mysqli_stmt_errno($stmt_1));
die("A. Prepare failed!");
}
mysqli_stmt_close($stmt_1);
$total_pages = ceil($row_count/$limit);
$offset = (($page * $limit) - $limit);
$query_2 = "SELECT id,$query_2)) //LINE 82:
{
echo __LINE__; echo "<br>";//LINE 84: THIS LINE GETS TRIGGERED AND SO PREPARED STATEMENT DID NOT FAIL AT LINE 82!
mysqli_stmt_bind_param($stmt_2,MYSQLI_ASSOC))//LINE 96: THIS LINE IS LIKE AN 'ELSE'. IT DOES NOT TRIGGER. NEITHER THE LINE IN THE 'IF'. WHY ?
{
echo __LINE__; echo "<br>";
echo "LIMIT: $limit<br>";
echo "ROW COUNT: $row_count<br><br>";
echo "TOTAL PAGES: $total_pages<br><br>";
//Retrieve Values.
$id = $row["id"];
$page_url = $row["page_url"];
$link_anchor_text = $row["link_anchor_text"];
$page_description = $row["page_description"];
$keyphrases = $row["keyphrases"];
$keywords = $row["keywords"];
echo "Id: $id<br>";
echo "Page Url: $page_url<br>";
echo "Link Anchor Text: $link_anchor_text<br>";
echo "Page Description: $page_description<br>";
echo "Keyphrases: $keyphrases<br>";
echo "Keywords: $keywords<br>";
echo "<br>";
echo "<br>";
}
}
else
{
echo __LINE__; echo "<br>";//LINE 124
printf("Error: %s.\n",mysqli_stmt_errno($stmt_2));
die("B. Prepare failed!");
}
mysqli_stmt_close($stmt_2);
//PAGINATION SECTION STARTS FROM HERE:
//WHY PAGINATION SECTION FAILS TO DISPLAY ? ONLY A LINK TO PAGE 1 IS SHOWN WHEN YOU LOAD PAGE 1. NO LINK TO PAGE 2. THERE ARE 2 MATCHING ROWS. SCRIPT FIXED TO SHOW 1 ROW PER PAGE. AND SO,2 PAGES SHOULD DISPLAY ALL RESULTS. HENCE,PAGINATION SECTION SHOULD SHOW A LINK TO PAGE 2 FROM PAGE 1. BUT PAGE 1 ONLY LINKS TO ITSELF IN THE PAGINATION SECTION.
if($page>$total_pages) //If Page Number is greater than Total Pages,show only a link to FINAL PAGE.
{
echo "?><a href=\"pagination_test.php?limit=$limit&page=$total_pages\">";?><?php echo "<b> Final Page </b>";?></a><?php
}
else
{
$i = 1;
while($i<=$total_pages)
{
if($i<$total_pages)
{
echo "<a href=\"pagination_test.php?keywords=$keywords&limit=$limit&page=$i\">";?><?php echo " $i ";?></a><?php
}
elseif($i==$page) //Bold the Current Page Number.
{
echo "<a href=\"pagination_test.php?keywords=$keywords&limit=$limit&page=$i\">";?><?php echo "<b> $i </b>";?></a><?php
}
$i++;
}
}
?>
再次更新: 问题:Url“&page =”中缺少。
看这个html:
<form method='GET' action="<?php echo $_SERVER['PHP_SELF'];?>?keywords=$keywords&limit=$limit&page=1">
如您所见,单击表单按钮后,您应该被发送到: pagination_test_2.php?keywords = $ keywords&limit = $ limit&page = 1“
但是你猜怎么着?您被发送到: ?keywords = $ keywords&limit = $ limit&search = search“
URL中“&page = 1”部分在哪里? 另外,如何不将“ search = search”添加到网址中?
<?php
require 'conn.php';
require 'error_reporting.php';
?>
<!DOCTYPE HTML">
<html>
<head>
<meta name="viewport" content="width-device=width,initial-scale=1">
</head>
<body>
<form method='GET' action="<?php echo $_SERVER['PHP_SELF'];?>?keywords=$keywords&limit=$limit&page=1">
<label for="keywords">Keywords:*</label>
<input type="text" name="keywords" id="keywords" placeholder="Input Keywords" required>
<br>
<label for="limit">Results per Page</label>
<select name="limit" id="limit">
<option value="1">1</option>
<option value="10">10</option>
<option value="25">25</option>
<option value="50">50</option>
<option value="100">100</option>
</select>
<br>
<button name="search" id="search" value=" ">Search</button><br>
<button type="submit" name="search" id="search" value="search">Search</button>
<br>
<input type="reset">
<br>
</form>
<?php
echo __LINE__; echo "<br>";
if(!ISSET($_GET['keywords']))
{
echo __LINE__; echo "<br>";
die("Type your keywords");
}
$keywords = $_GET['keywords'];
//Check if the PAGE NUMBER is specified or not and if it's a numer or not. If not,$query_1))
{
echo __LINE__; echo "<br>";
mysqli_stmt_bind_param($stmt_1,$row_count);
mysqli_stmt_fetch($stmt_1);
echo "ROW COUNT: $row_count<br><br>";
}
else
{
echo __LINE__; echo "<br>";
printf("Error: %s.\n",$limit";
$stmt_2 = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt_2,$query_2))
{
echo __LINE__; echo "<br>";
mysqli_stmt_bind_param($stmt_2,$keywords);
mysqli_stmt_execute($stmt_2);
$result_2 = mysqli_stmt_get_result($stmt_2);
if(!$result_2)
{
echo __LINE__; echo "<br>";
die("Fetching Error");
}
while($row = mysqli_fetch_array($result_2,MYSQLI_ASSOC))
{
echo __LINE__; echo "<br>";
echo "LIMIT: $limit<br>";
echo "ROW COUNT: $row_count<br><br>";
echo "TOTAL PAGES: $total_pages<br><br>";
//Retrieve Values.
$id = $row["id"];
$page_url = $row["page_url"];
$link_anchor_text = $row["link_anchor_text"];
$page_description = $row["page_description"];
$keyphrases = $row["keyphrases"];
$keywords = $row["keywords"];
echo "Id: $id<br>";
echo "Page Url: $page_url<br>";
echo "Link Anchor Text: $link_anchor_text<br>";
echo "Page Description: $page_description<br>";
echo "Keyphrases: $keyphrases<br>";
echo "Keywords: $keywords<br>";
echo "<br>";
echo "<br>";
}
}
else
{
echo __LINE__; echo "<br>";
printf("Error: %s.\n",mysqli_stmt_errno($stmt_2));
die("B. Prepare failed!");
}
mysqli_stmt_close($stmt_2);
if($page>$total_pages) //If Page Number is greater than Total Pages,show only a link to FINAL PAGE.
{
echo "<a href=\"pagination_test.php?keywords=$keywords&limit=$limit&page=$total_pages\">"; echo "<b> Final Page </b>";?></a><?php
}
else
{
$i = 1;
while($i<=$total_pages)
{
if($i<$total_pages && $i!=$page)
{
echo "<a href=\"pagination_test_2.php?keywords=$keywords&limit=$limit&page=$i\">"; echo " $i ";?></a><?php
}
elseif($i==$page) //Bold the Current Page Number.
{
echo "<a href=\"pagination_test_2.php?keywords=$keywords&limit=$limit&page=$i\">"; echo "<b> $i </b>";?></a><?php
}
else
{
echo "<a href=\"pagination_test_2.php?keywords=$keywords&limit=$limit&page=$i\">"; echo " $i ";?></a><?php
}
$i++;
}
}
?>
错别字位于分页部分的网址上。现在,我修复了它们,因此分页部分工作正常。 有人可以友好地签出我的代码,并让我知道如果不是真的需要淘汰哪些行。
解决方法
我认为这两行都没有达到,因为该语句有效,但返回零行。在$query_2 = ...
之前,您可以在$row_count
上添加检查以查看是否有任何行吗?
要回答第二个问题,$total_pages = ceil($row_count/$limit);
是正确的答案,因为mysqli_stmt_bind_result
返回一个布尔值,因此$result_1
是对还是错。
我个人喜欢使用面向对象的样式,而不喜欢您使用的过程样式,因为我发现它更易于阅读,但选择是您自己的。
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