Jer*_*emy 3 php mysql while-loop
我测试了循环嵌套的While语句,以便:
$count1 = 0;
while ($count1 < 3) {
$count1++;
$count2 = 0;
echo "count1: ".$count1."<br />";
while ($count2 < 3) {
$count2++;
echo "count2: ".$count2."<br />";
}
}
Run Code Online (Sandbox Code Playgroud)
效果完美(每次循环三遍),结果如下:
count1: 1
count2: 1
count2: 2
count2: 3
count1: 2
count2: 1
count2: 2
count2: 3
count1: 3
count2: 1
count2: 2
count2: 3
Run Code Online (Sandbox Code Playgroud)
然后我尝试使用mysql_fetch_assoc($ ContactsInterests是两行关联数组,而$ LatestNews有50行)循环进行相同的操作,即
$CI_count = 0;
while ($CI_Row = mysql_fetch_assoc($ContactsInterests)) { //loop thru interests
$CI_count++;
$LN_count = 0;
echo "CI_count: ".$CI_count."<br />";
while ($LN_Row = mysql_fetch_assoc($LatestNews)) { //loop thru news
$LN_count++;
echo "LN_count: ".$LN_count."<br />";
}
}
Run Code Online (Sandbox Code Playgroud)
结果是:
CI_count: 1
LN_count: 1
LN_count: 2
...
LN_count: 50
LN_count: 51
CI_count: 2
Run Code Online (Sandbox Code Playgroud)
但是LN_count的第二次迭代在哪里?我不明白为什么LN_count没有第二次增加。
帮助表示赞赏。
mysql_fetch_assoc对“ mysql结果”类型进行迭代。寻求每个提取的索引。您必须使用mysql_data_seek转到第一个结果,例如;
<?php
$CI_count = 0;
while ($CI_Row = mysql_fetch_assoc($ContactsInterests)) { //loop thru interests
$CI_count++;
$LN_count = 0;
echo "CI_count: ".$CI_count."<br />";
mysql_data_seek($LatestNews,0);
while ($LN_Row = mysql_fetch_assoc($LatestNews)) { //loop thru news
$LN_count++;
echo "LN_count: ".$LN_count."<br />";
}
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
3601 次 |
| 最近记录: |